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PRACTICE / Heat capacity

Find the energy taken by the cup

Two heater trials each raise a sample and the same cup by 10 K from the same initial temperature. A 0.150 kg sample requires 2100 J; a 0.300 kg sample of the same material requires 3600 J. The heater delivers all its energy to sample and cup, without losses.

Equal temperature rises in the same cup
Sample mass (kg)Delivered heat (J)Rise (K)
0.150210010
0.300360010

What is the cup heat capacity?

Organize the givens & assumptions
  • Masses = 0.150 and 0.300 kg
  • Inputs = 2100 and 3600 J
  • Same 10 K rise
  • Same unknown sample specific heat

Ideal insulated calorimetry at ordinary atmospheric pressure, no evaporation or other work, equilibrium final state and stated constant model capacities unless a temperature dependence is specified. Vessel heat capacity is negligible except where it is an explicit unknown. Pure ice/water melts at 0°C; no supercooling.

  1. A60 J/K60\ \mathrm{J/K}
  2. B210 J/K210\ \mathrm{J/K}
  3. C150 J/K150\ \mathrm{J/K}
  4. D600 J/K600\ \mathrm{J/K}

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