A little hot water can warm a larger amount of cool water, but the final temperature is not generally the arithmetic mean. The sensible way to predict it is to follow energy from the parts that cool to the parts that warm.
We will develop that accounting for calorimetry, then study how geometry controls conduction and how a temperature-dependent loss produces an exponential cooling curve.
What you will learn to do
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BEFORE YOU READ
Bring back one idea
Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.
Why can two objects at the same temperature contain different amounts of thermal energy relative to a chosen reference?
Compare your reasoning
Temperature is not an amount of stored energy. Mass, material, phase and the chosen reference matter. Heat capacity tells us the energy needed for a specified temperature change in a stated process.
Revisit the explanation →01 / Heat balances, conduction and cooling
Equal temperature is an equilibrium statement
A thermometer works because two bodies in thermal equilibrium have the same temperature. If each separately equilibrates with the same thermometer, they are in thermal equilibrium with one another. This is the operational idea behind the zeroth law: temperature makes thermal equilibrium comparable across different objects.
Equal temperature does not mean equal internal energy or equal heat capacity. A cup and a swimming pool can share a temperature while requiring very different energies for the same temperature rise. When bodies exchange heat, the final temperature follows an energy balance with their capacities, not an unweighted average of their initial temperatures.
A steady temperature also need not mean equilibrium. A powered heater can keep a plate at a constant temperature while heat continuously flows into a cooler room. Equilibrium means no net driving temperature differences among the bodies under consideration; a steady state merely means the measured properties stop changing with time.
WORKED EXAMPLE
Check a thermometer story with an energy balance
An insulated pair of bodies has heat capacities 120 J/K and 80 J/K, initially at 40°C and 10°C. Their capacities are constant, and no phase change occurs. Find their equilibrium temperature.
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STOP & THINK
A powered plate remains at 60°C while its room remains at 20°C. Is the plate necessarily in thermal equilibrium with the room?
- Yes, because its temperature is constant
- No; steady heat input can balance heat loss
- Yes, if its heat capacity is constant
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Focus: Equal temperature is an equilibrium statement Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
02 / Heat balances, conduction and cooling
Heat capacity links an energy transfer to a temperature change
If a body stays in one phase and its heat capacity is approximately constant, the energy required for a temperature increase is Q = CΔT = mcΔT. Heat capacity C belongs to the whole body and has units J/K. Specific heat capacity c is per unit mass and has units J/(kg·K).
A temperature interval has the same numerical value in kelvins and Celsius degrees, so either can be used for ΔT in this relation. This is different from the absolute-temperature ratios in gas laws or the fourth powers in radiation formulas, where kelvins are required.
The relation assumes that the stated heat capacity describes the process. For solids and liquids in ordinary calorimetry, volume changes and associated work are often negligible. For gases, constant-pressure and constant-volume heat capacities differ, so the process constraint matters.
If heat capacity varies appreciably with temperature, replace CΔT by an integral of C(T)dT. If a phase change occurs, include the latent-energy transfer separately. A single straight temperature-rise formula cannot cross every material change without adjustment.
Use this constant-capacity form within one phase and under the specified process conditions.
STOP & THINK
Two bodies have the same specific heat capacity and receive the same energy with no phase change. One has twice the mass. Its temperature rise is what?
- Twice as large
- Half as large
- The same
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Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Focus: Heat capacity links an energy transfer to a temperature change Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Equations in this section (LaTeX): Q=mc\Delta T=C\Delta T Meaning: Use this constant-capacity form within one phase and under the specified process conditions. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
03 / Heat balances, conduction and cooling
Use measurements to separate sample and container
A calorimeter is part of the energy balance even when its walls are well insulated. Insulation describes exchange with the environment. Heat capacity describes how much energy the cup, stirrer and sensor take up as their temperature changes. These are different properties.
If the same temperature rise is produced with two different sample masses in the same apparatus, the change in supplied energy isolates the sample contribution. The cup contribution stays the same and cancels in a difference. This is the same experimental habit used when a scale has an unknown zero: learn from a controlled change before interpreting an absolute reading.
WORKED EXAMPLE
Measure a material without pretending the cup is massless
A lossless heater warms a sample and its cup by 10 K. A 0.100 kg sample requires 1300 J; a 0.200 kg sample of the same material in the same cup requires 2100 J for the same rise. Each trial starts at the same temperature. Find the sample specific heat c and the cup heat capacity C.
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For a temperature-dependent heat capacity, the temperature path matters. Divide the path into small intervals, add mc(T)dT, and include a separate latent term if the path crosses a phase transition. An endpoint specific heat is not automatically the average over the interval.
WORKED EXAMPLE
Integrate a changing heat capacity
A 0.50 kg single-phase sample has c(θ) = 400 + 2θ J/(kg·K), where θ is the numerical rise in kelvins from its initial temperature. Find the heat needed for a 30 K rise, with negligible work and losses.
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STOP & THINK
An insulated cup has a nonzero heat capacity. Which statement follows?
- It cannot change temperature
- It can exchange energy with its contents while the combined system remains insulated
- It must have zero mass
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Focus: Use measurements to separate sample and container Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
04 / Heat balances, conduction and cooling
An insulated mixture requires an energy balance for every participating body
When hot and cold bodies are placed together in an insulated enclosure, energy transfers internally until they reach a common equilibrium temperature, provided no phase change or reaction complicates the model. The total energy change of the chosen combined system is zero.
For constant heat capacities, sum C_i(T_f − T_i) over all participating bodies and set the result to zero. Hot bodies contribute negative changes and cold bodies positive ones. The resulting temperature is a heat-capacity-weighted average, not necessarily the ordinary average of the initial temperatures.
The container, thermometer and stirrer can exchange energy too. If their heat capacities are significant, include them with their own initial temperatures. Calling a vessel insulated only limits exchange with the outside; it does not give the vessel zero heat capacity.
The final temperature must lie between the initial temperatures for this no-work, no-phase-change model. A result outside that range signals a missing energy source, an incorrect sign or an invalid assumption. Evaporation, heat leakage and incomplete equilibration are common experimental departures.
This weighted-average result assumes an insulated collection with constant positive heat capacities, no phase changes and negligible work.
WORKED EXAMPLE
Include the calorimeter rather than averaging only the liquids
Mix 0.30 kg of water at 80°C with 0.20 kg of water at 20°C in an insulated calorimeter initially at 20°C. The calorimeter heat capacity is 420 J/K. Use c_water = 4200 J/(kg·K). Find the equilibrium temperature.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling
Focus: An insulated mixture requires an energy balance for every participating body
Learning goals:
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Equations in this section (LaTeX):
\sum_i C_i(T_f-T_i)=0,\qquad T_f=\frac{\sum_i C_iT_i}{\sum_i C_i}
Meaning: This weighted-average result assumes an insulated collection with constant positive heat capacities, no phase changes and negligible work.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.05 / Heat balances, conduction and cooling
A phase change needs an energy budget before a final temperature
During an equilibrium phase change of a pure substance at the relevant fixed pressure, energy can change the phase fraction without changing temperature. The required transfer is mL, where L is the specific latent heat under those conditions.
For ice initially below its melting temperature, first warm the ice to the melting point, then melt some or all of it, and only then warm the liquid that has formed. Each stage has a different energy expression. A negative final temperature obtained from an assumed all-liquid state often means that the melting assumption was wrong.
A safe method is to compare the energy available from the warmer bodies down to the melting temperature with the energy needed to warm and melt the cold material. If enough energy is available, continue with a liquid-temperature balance. If not, solve for how much melts and leave the final mixture at the melting point.
The examples below use an idealized water–ice system near ordinary atmospheric pressure with melting temperature 0°C, negligible vessel heat capacity and no heat loss unless stated. Values of heat capacities and latent heat are supplied explicitly rather than treated as universal exact constants.
Latent heat L has units J/kg. It applies to the specified phase transition and pressure, not to an arbitrary temperature change.
WORKED EXAMPLE
Follow warming, melting and final warming in order
Mix 0.050 kg of ice at −10°C with 0.20 kg of water at 40°C in an ideal insulated vessel. Use c_ice = 2100 J/(kg·K), c_water = 4200 J/(kg·K) and L_f = 336000 J/kg. Find the final state.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling
Focus: A phase change needs an energy budget before a final temperature
Learning goals:
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Equations in this section (LaTeX):
Q_{\mathrm{phase}}=mL
Meaning: Latent heat L has units J/kg. It applies to the specified phase transition and pressure, not to an arbitrary temperature change.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.06 / Heat balances, conduction and cooling
If melting is incomplete, solve for mass rather than forcing a temperature
If the available energy is insufficient to melt all ice already at 0°C, liquid water and ice coexist at 0°C in the ideal equilibrium model. The unknown is the melted mass, found by dividing available energy by latent heat.
Do not put the entire original ice mass into a liquid heat-capacity term unless it actually melts. Likewise, do not assume every mixing problem must end with one uniform phase. The final temperature and phase composition are both parts of the answer.
At the boundary where the energy just suffices to melt all the ice, the final state is all liquid at the melting temperature. Any additional energy then raises the temperature. Checking this boundary gives a useful consistency test between partial-melting and complete-melting calculations.
WORKED EXAMPLE
Determine the remaining ice when energy is insufficient
Mix 0.10 kg of ice at 0°C with 0.20 kg of water at 30°C in an ideal insulated vessel. Use c_water = 4200 J/(kg·K) and L_f = 336000 J/kg. Find the final temperature and remaining ice mass.
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STOP & THINK
A calculation assumes all ice melts but gives a final all-liquid temperature below 0°C in the stated equilibrium water–ice model. What should you do?
- Accept the negative result without checking
- Revisit the assumed final phase and compare latent-energy budgets
- Change the sign of the latent heat arbitrarily
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Focus: If melting is incomplete, solve for mass rather than forcing a temperature Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
07 / Heat balances, conduction and cooling
Choose the final phase using a common energy reference
A final temperature alone does not describe an ice–water mixture. At the melting point, two mixtures can have the same temperature and different amounts of liquid. Their energy contents differ because melting requires latent energy. The phase fraction is part of the state you must determine.
Choose ice at 0°C as the zero of an energy ledger. One kilogram of colder ice then has value cᵢT, where T is its signed Celsius temperature relative to 0°C. One kilogram of liquid water has value Lf + c𝓌T: it first needs the melting energy and then any warming energy. Do not compare ice and water using mcT alone with two unrelated zeroes.
At fixed atmospheric pressure this ledger uses enthalpy H, which includes pressure–volume work. The heat capacities and fusion energy below describe that constant-pressure process. For an insulated mixture with no other work, no evaporation and negligible vessel capacity, the total H is unchanged.
Let M be the total mass. If the available H is below zero, even all the mass as ice at 0°C would require more energy than is available. If H is between zero and MLf, the final state is at 0°C with some ice and some water. If H exceeds MLf, enough energy exists to melt everything and warm the liquid. This branch test comes before solving a temperature.
Use a shared reference: ice at 0°C. The final mass relation applies only in the coexistence interval, including its pure-phase endpoints.
Explain this equation
Read it in words
Add the initial ice and water enthalpies, then use the melting interval to determine the final liquid mass.
Symbols and units
- Total enthalpy relative to the same mass of ice at 0°C.J.
- Initial ice and liquid-water masses.kg.
- Stated constant specific heat capacities of ice and water.J/(kg·K).
- Initial temperatures measured as signed intervals from 0°C.Numerically in °C; their interval sizes are in K.
- Positive specific latent heat of fusion at the stated pressure.J/kg.
- Conserved total mass mᵢ + m𝓌.kg.
- Final temperature on the melting-point branch.°C.
- Final liquid-water mass; the remaining M − m𝓌,f is ice.kg.
Signs and operations
Cold ice contributes a negative term. Initial water has a positive latent term even at 0°C. A reference value may be negative; it is not an assertion of negative absolute internal energy.
When it applies
Pure water/ice in equilibrium at ordinary atmospheric pressure, melting point taken as 0°C, constant specified capacities, negligible vessel capacity and no environmental heat, evaporation or other work. Supercooling and freezing kinetics are excluded.
Make it concrete
For 0.10 kg ice at −20°C and 0.15 kg water at 20°C with cᵢ = 2100, c𝓌 = 4200 J/(kg·K) and Lf = 334000 J/kg, H = 58500 J. This is between 0 and MLf = 83500 J, so about 0.17515 kg is liquid at 0°C.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Equation (LaTeX): \begin{aligned}H&=m_i c_iT_i+m_w(L_f+c_wT_w)\\ T_f&=0\quad\text{if }0\le H\le ML_f\\ m_{w,f}&=H/L_f\quad\text{on this branch}\end{aligned}
Read it as: Add the initial ice and water enthalpies, then use the melting interval to determine the final liquid mass.
Symbols:
H: Total enthalpy relative to the same mass of ice at 0°C.; J.
m_i, m_w: Initial ice and liquid-water masses.; kg.
c_i, c_w: Stated constant specific heat capacities of ice and water.; J/(kg·K).
T_i, T_w: Initial temperatures measured as signed intervals from 0°C.; Numerically in °C; their interval sizes are in K.
L_f: Positive specific latent heat of fusion at the stated pressure.; J/kg.
M: Conserved total mass mᵢ + m𝓌.; kg.
T_f: Final temperature on the melting-point branch.; °C.
m_{w,f}: Final liquid-water mass; the remaining M − m𝓌,f is ice.; kg.
Signs and operations: Cold ice contributes a negative term. Initial water has a positive latent term even at 0°C. A reference value may be negative; it is not an assertion of negative absolute internal energy.
Conditions: Pure water/ice in equilibrium at ordinary atmospheric pressure, melting point taken as 0°C, constant specified capacities, negligible vessel capacity and no environmental heat, evaporation or other work. Supercooling and freezing kinetics are excluded.
Help me explain the equation in words, predict how changing a quantity affects the result, and apply it to a small numerical example.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.WORKED EXAMPLE
First warm the ice, then count the part that melts
Mix 0.100 kg of ice at −20°C with 0.150 kg of water at 20°C in the ideal insulated, negligible-capacity vessel. Use cᵢ = 2100 J/(kg·K), c𝓌 = 4200 J/(kg·K) and Lf = 334 kJ/kg. Find the equilibrium temperature and remaining ice mass.
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STOP & THINK
While ice and water coexist at 0°C in the stated equilibrium model, a small extra heat input is supplied. What can increase before the temperature rises?
- The liquid mass
- The fusion energy per kilogram by the added heat
- Nothing can change at fixed temperature
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling
Focus: Choose the final phase using a common energy reference
Learning goals:
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Equations in this section (LaTeX):
\begin{aligned}H&=m_i c_iT_i+m_w(L_f+c_wT_w)\\ T_f&=0\quad\text{if }0\le H\le ML_f\\ m_{w,f}&=H/L_f\quad\text{on this branch}\end{aligned}
Meaning: Use a shared reference: ice at 0°C. The final mass relation applies only in the coexistence interval, including its pure-phase endpoints.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.08 / Heat balances, conduction and cooling
A cold mixture can freeze water instead of melting ice
Do not force every ice–water question into “some ice melts.” Very cold ice may need more warming energy than the water can provide by cooling alone. Freezing some of the liquid releases latent energy. If even freezing all of it is insufficient to bring everything to 0°C, the final state is entirely ice below 0°C.
At the other extreme, if enough initial energy exists to melt everything, only the remaining energy warms the combined liquid mass. Use the final total mass in that last heat-capacity term. Using only the melted ice mass would ignore the water that was already present.
A negative temperature obtained after assuming all-liquid water is a failed branch assumption in this equilibrium model. Likewise, a calculated liquid mass outside zero to total mass means the coexistence branch was chosen incorrectly. These checks diagnose the model choice before you blame the arithmetic.
The first branch contains only ice and the second only liquid. Temperatures are relative to 0°C; a liquid result reaching boiling would require an additional vapour branch not included here.
WORKED EXAMPLE
When all the water freezes
Combine 0.300 kg ice at −40°C with 0.050 kg water at 0°C in an ideal insulated vessel of negligible heat capacity. Use cᵢ = 2100 J/(kg·K), c𝓌 = 4200 J/(kg·K) and Lf = 334 kJ/kg as constant model values. Find the final equilibrium state.
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WORKED EXAMPLE
After melting, warm the entire liquid mass
Mix 0.050 kg ice at −10°C with 0.200 kg water at 40°C under the same ideal model. Use cᵢ = 2100 J/(kg·K), c𝓌 = 4200 J/(kg·K) and Lf = 334 kJ/kg. Find the final state.
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INTERACTIVE LAB
Explore the final phase before the temperature
Predict whether the final state is all ice, a mixture at 0°C or all liquid. Change one initial mass or temperature, then compare the common-reference energy with the two phase boundaries. The bars show mass fractions, not occupied volumes or an animation of mixing.
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STOP & THINK
Under the worked all-liquid case above, what final temperature in °C follows from a 15850 J surplus and total heat capacity 1050 J/K?
Answer in °C.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling
Focus: A cold mixture can freeze water instead of melting ice
Learning goals:
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Equations in this section (LaTeX):
\begin{aligned}T_f&=\frac{H}{Mc_i}&& (H<0)\\ T_f&=\frac{H-ML_f}{Mc_w}&& (H>ML_f)\end{aligned}
Meaning: The first branch contains only ice and the second only liquid. Temperatures are relative to 0°C; a liquid result reaching boiling would require an additional vapour branch not included here.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.09 / Heat balances, conduction and cooling
Conduction transfers energy down a temperature gradient
Conduction transfers thermal energy through microscopic interactions without requiring bulk transport of the material. For steady one-dimensional flow through a uniform slab with insulated sides, constant thermal conductivity k, area A and thickness L, the heat-transfer rate is kA(T_h − T_c)/L.
The rate has units watts; the energy transferred during a time interval is the integral of that rate over time. A steady rate multiplied by time gives energy, but a changing rate cannot always be replaced by its initial value for the entire process.
The local form is Q̇ = −kA dT/dx. The minus sign means energy flows toward lower temperature. In a uniform slab at steady state with no internal heat generation and constant k, the temperature profile is linear.
Steady does not mean equal temperature everywhere. It means the temperature at each position no longer changes with time. A maintained temperature gradient can support a constant flow of energy through the slab without accumulating energy in it.
The second expression uses a uniform slab, constant k, steady one-dimensional conduction and negligible side losses.
WORKED EXAMPLE
Distinguish a heat-transfer rate from an energy amount
A slab has k = 0.50 W/(m·K), area 0.020 m² and thickness 0.010 m. Its faces are maintained at 70°C and 30°C, with insulated sides. Find the steady heat-transfer rate and energy transferred in 2 minutes.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling
Focus: Conduction transfers energy down a temperature gradient
Learning goals:
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Equations in this section (LaTeX):
\dot Q=-kA\frac{dT}{dx},\qquad \dot Q=\frac{kA(T_h-T_c)}{L}
Meaning: The second expression uses a uniform slab, constant k, steady one-dimensional conduction and negligible side losses.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.10 / Heat balances, conduction and cooling
Series layers share a heat current; parallel paths share a temperature difference
Define a slab’s thermal resistance as R_th = L/(kA), in K/W. Then the steady heat current is ΔT/R_th. This definition includes the area; some engineering R-values are instead area-normalized, so compare units before using a quoted resistance.
For slabs in series with no heat generation or side loss, the same steady heat current passes through every layer. Temperature drops add, so their thermal resistances add. The interface temperature is determined by the drop across either layer.
For parallel paths connecting the same two reservoirs, each path has the same temperature difference and their heat currents add. Conductances 1/R_th therefore add. A thin highly conducting path can substantially increase the total heat leak even if most of the wall is well insulated.
Contact resistance can create an additional temperature drop at an interface. If it is neglected, the ideal model has a common interface temperature. Treat a supplied contact resistance as another series resistance rather than hiding it in a bulk conductivity.
These network rules require the stated steady-flow connections and no unaccounted heat sources or side losses.
WORKED EXAMPLE
Find both heat current and the interface temperature
Two slabs share area 0.010 m² and are placed in series between 80°C and 20°C reservoirs. From the hot side, slab 1 has L₁ = 0.020 m and k₁ = 0.20 W/(m·K); slab 2 has L₂ = 0.040 m and k₂ = 0.80 W/(m·K). Neglect contact resistance and side losses. Find the steady heat current and interface temperature.
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STOP & THINK
Two thermal paths of resistances 10 K/W and 20 K/W connect reservoirs differing by 60 K. What is the total steady heat-transfer rate?
Answer in W.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling
Focus: Series layers share a heat current; parallel paths share a temperature difference
Learning goals:
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Equations in this section (LaTeX):
R_{\mathrm{th}}=\frac{L}{kA},\qquad R_{\mathrm{series}}=\sum_iR_i,\qquad \frac1{R_{\mathrm{parallel}}}=\sum_i\frac1{R_i}
Meaning: These network rules require the stated steady-flow connections and no unaccounted heat sources or side losses.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.11 / Heat balances, conduction and cooling
Newton cooling balances stored energy against a temperature-dependent loss
A small body with nearly uniform internal temperature can sometimes be represented by one temperature T(t) and a constant heat capacity C. Suppose its net heat-loss rate is approximately G(T − T_a), where T_a is a constant ambient temperature and G is an effective thermal conductance.
With no applied heating and negligible work, energy balance gives C dT/dt = −G(T − T_a). Define α = G/C, in s⁻¹. The temperature excess θ = T − T_a then obeys dθ/dt = −αθ, giving an exponential approach to ambient temperature.
This lumped model requires internal temperature differences to be small and the effective loss coefficient to remain approximately constant. It may describe a limited range of convection or a linearized radiative loss, but it is not a universal cooling law for every temperature and geometry.
As the temperature difference shrinks, the cooling rate shrinks too. Equal decreases in absolute temperature therefore do not generally take equal times. Instead, equal fractional reductions of the temperature excess take equal times.
Use a constant ambient temperature, constant heat capacity and approximately linear loss rate. Time units must match those in α.
Cool by a fraction of the excess temperature
- Body
- 80
- Ambient
- 20
STOP & THINK
Under the constant-ambient Newton-cooling model, what happens to a body initially hotter than its surroundings?
- It reaches ambient in a finite time and keeps cooling below it
- Its temperature excess decreases exponentially toward zero
- It loses the same number of degrees every second
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling
Focus: Newton cooling balances stored energy against a temperature-dependent loss
Learning goals:
- Distinguish heat capacity from specific heat and heat-transfer rate.
- Include calorimeter energy in a mixing balance.
- Test phase assumptions before solving a final temperature.
- Calculate steady conduction through single, series and parallel paths.
- Use interface temperatures to check a common heat current.
- Derive Newton cooling from an energy-rate balance.
- Infer cooling behaviour from temperature excess and separate steady heating from equilibrium.
- Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference.
Equations in this section (LaTeX):
\frac{dT}{dt}=-\alpha(T-T_a),\qquad T(t)=T_a+(T_i-T_a)e^{-\alpha t},\qquad\alpha=\frac GC
Meaning: Use a constant ambient temperature, constant heat capacity and approximately linear loss rate. Time units must match those in α.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.12 / Heat balances, conduction and cooling
Infer a cooling constant from temperature excesses
Given two temperatures separated by a known time in a constant ambient, form the ratio (T₂ − T_a)/(T₁ − T_a). Its logarithm equals −αΔt. Subtract the ambient before taking a ratio; raw Celsius temperatures do not have the required exponential relation.
Once α is known, the time to cool between any two temperatures above ambient is (1/α) ln[(T₁ − T_a)/(T₂ − T_a)]. The denominator temperature excess must remain positive in this hotter-than-ambient cooling case.
A plot of ln(T − T_a) against time should be straight if the model holds, with slope −α. Curvature can indicate changing loss mechanisms, a changing ambient or significant internal temperature gradients. A good straight fit supports the model over that data range; it does not guarantee it at much higher temperatures.
Using an average temperature in a finite-interval rate formula is an approximation. The exponential solution is available directly for the constant-α model, so it avoids treating a long cooling interval as if the instantaneous rate were constant.
WORKED EXAMPLE
Predict a later temperature from one measured fractional drop
A body cools from 80°C to 50°C in 6 minutes in a room held at 20°C. Assume a constant-α Newton-cooling model. Find its temperature 12 minutes after the initial 80°C reading.
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WORKED EXAMPLE
Compare equal degree drops at different excess temperatures
In the same 20°C ambient, compare the time to cool from 80°C to 60°C with the time to cool from 60°C to 40°C under a fixed cooling constant α.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Focus: Infer a cooling constant from temperature excesses Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
13 / Heat balances, conduction and cooling
A steady final temperature can coexist with continuous heat flow
If a heater supplies constant power P while the same linear loss mechanism removes G(T − T_a), the energy balance becomes C dT/dt = P − G(T − T_a). At the steady temperature, storage stops changing, so P = G(T_steady − T_a).
The body need not reach thermal equilibrium with the room. It reaches a steady state with heat continuously entering from the heater and leaving to the surroundings. This distinction parallels a conducting slab whose temperatures stay fixed while energy flows through it.
The approach to the new steady temperature remains exponential in the constant-parameter lumped model, with time constant C/G. Increasing heat capacity slows the approach but does not change the final temperature for fixed P and G. Increasing heat-loss conductance lowers the final temperature excess and speeds the response.
In an experiment, include the container’s heat capacity in C and check that stirring or internal conduction keeps the sample nearly uniform. Avoid using the model through a phase transition, where temperature can remain nearly fixed while the stored energy changes substantially.
WORKED EXAMPLE
Separate the steady temperature from the response time
A uniformly heated body has heat capacity 200 J/K and loss conductance 10 W/K to a 20°C environment. A heater supplies 100 W continuously. Under the constant-parameter model, find the steady temperature and time constant.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Focus: A steady final temperature can coexist with continuous heat flow Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
14 / Heat balances, conduction and cooling
Moving fluid can carry thermal energy
In conduction, energy passes through matter without requiring bulk transport of the material. Convection couples heat transfer to moving fluid. A parcel of warmed fluid can carry energy away from a surface; another parcel replaces it. Heat still has to cross the thin fluid layer next to the surface, so flow and local conduction act together.
Natural convection can be driven by buoyancy when temperature-dependent density differences make a fluid arrangement unstable. Forced convection uses a pump or fan. Warm fluid does not inevitably rise in every setting: gravity, boundaries, density behaviour and the surrounding flow matter.
An empirical surface law is heat-transfer rate = hA(Ts − Tfluid), where h has units W/(m² K). The coefficient summarizes a flow and geometry; it is not a universal material constant. For a steady stream with negligible kinetic/potential changes and no phase transition, the energy gained per second is mass-flow rate times specific heat times temperature rise.
WORKED EXAMPLE
Connect heater power to a moving stream
A steady water stream gains 420 W from a heater. Its mass-flow rate is 0.010 kg/s and specific heat is 4200 J/(kg K). Neglect losses and changes of kinetic and gravitational energy. Find its temperature rise.
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STOP & THINK
A fan changes the flow past a warm object. Must the convection coefficient h remain the same because the object material is unchanged?
- Yes
- No; h depends on the flow as well as geometry
- Only the object mass determines h
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Heat balances, conduction and cooling Focus: Moving fluid can carry thermal energy Learning goals: - Distinguish heat capacity from specific heat and heat-transfer rate. - Include calorimeter energy in a mixing balance. - Test phase assumptions before solving a final temperature. - Calculate steady conduction through single, series and parallel paths. - Use interface temperatures to check a common heat current. - Derive Newton cooling from an energy-rate balance. - Infer cooling behaviour from temperature excess and separate steady heating from equilibrium. - Distinguish all-ice, coexistence and all-liquid outcomes from a shared phase-energy reference. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BRING THE IDEAS TOGETHER
What should stay with you
- Constant-capacity temperature changes and latent phase changes need separate energy terms.
- An insulated vessel can still have a nonzero heat capacity.
- Insufficient melting energy leads to a phase mixture at the transition temperature.
- Steady conduction can occur through a nonuniform temperature profile.
- Thermal resistances add in series; conductances add in parallel.
- Newton cooling is exponential in temperature excess above ambient.
- Equal fractional excess reductions take equal times; equal degree drops do not.
- A heated steady state can have continuous input and loss despite no temperature change.
RETRIEVE, THEN EXPLAIN
Try it without the examples
Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.
SELF CHECK 1
A material has latent heat 200000 J/kg. How much mass changes phase when 6000 J is used only for that transition?
Answer in kg.
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SELF CHECK 2
Two series thermal resistances are 4 K/W and 6 K/W. Their reservoirs differ by 50 K. What is the steady heat current?
Answer in W.
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SELF CHECK 3
A body starts at 80°C in a 20°C ambient. Its excess temperature halves every 6 minutes. What is its temperature after 18 minutes?
Answer in °C.
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SELF CHECK 4
A slab conducts heat steadily between maintained hot and cold faces. What does steady mean?
- All points have equal temperature
- The temperature at each point no longer changes with time
- No energy passes through the slab
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SELF CHECK 5
To infer a Newton-cooling constant in an ambient T_a, which ratio belongs inside the logarithm?
- T₂/T₁ using raw Celsius readings
- (T₂ − T_a)/(T₁ − T_a)
- The masses of the body at the two times
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APPLY THE CHAPTER
Take the ideas into a problem
Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.
Two portions of water reach equilibriumIntermediate · Free sample · Hints and guided reasoningHeat through a metal rodIntermediate · Free sample · Hints and guided reasoningTemperature equilibrium does not average equallyBeginner · Hints and guided reasoningA steady surface can keep transferring heatIntermediate · Hints and guided reasoningFind the energy taken by the cupIntermediate · Hints and guided reasoningUse the whole heat-capacity curveAdvanced · Hints and guided reasoningReserve energy for warming the iceIntermediate · Hints and guided reasoningFreezing can supply the missing warming energyIntermediate · Hints and guided reasoningReject an impossible coexistence massAdvanced · Hints and guided reasoningFind the boundary between a mixture and all liquidAdvanced · Hints and guided reasoningEnergy that changes phaseIntermediate · Hints and guided reasoningTwo stages for the same sampleIntermediate · Hints and guided reasoningTwo slabs share a temperature dropIntermediate · Hints and guided reasoningCooling slows near room temperatureIntermediate · Hints and guided reasoningHalve the excess twiceIntermediate · Hints and guided reasoningFind the hidden interface temperatureIntermediate · Hints and guided reasoningCOME BACK TO THE IDEA
What can you explain later?
Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.
Next study session
Can adding cold ice to water at 0°C increase the amount of ice instead of melting it?
Compare your reasoning
Yes. Freezing some water can release the latent energy needed to warm the colder ice. If even freezing all the water cannot reach 0°C, the final equilibrium is entirely ice below 0°C in this model.
Revisit the explanation →A few days later: change the situation
Two identical warm objects cool in the same surroundings, one with twice the initial temperature excess. Under Newton’s cooling model, compare their initial cooling rates.
Compare your reasoning
The object with twice the temperature excess initially cools twice as fast in magnitude. Both have the same fractional decay rate if the heat-transfer coefficient and heat capacity remain constant.
Revisit the explanation →Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.
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