Calorimetry with a container that absorbs heat
A student places a hypothetical metal specimen into water in an insulated calorimeter. All numerical thermal properties are supplied for this exercise rather than inferred from a material name. The specimen has mass 0.250 kg, specific heat capacity 400 J per kilogram per kelvin and initial temperature 100.0 degrees Celsius. The water mass is 0.100 kg and its specific heat capacity is 4200 J per kilogram per kelvin. The water and container begin together at 20.0 degrees Celsius. The container has heat capacity 20 J per kelvin. Assume there is no evaporation, no phase change, negligible heat exchange with the surroundings and complete thermal equilibrium at the final reading. A classmate ignores the container because it has no liquid inside its walls. The student must decide how that omission changes the predicted final temperature.
1. Calculate the equilibrium temperature including the container.
Read the worked answer
The metal heat capacity is 0.250 times 400 = 100 J/K. Water plus container have heat capacity 0.100 times 4200 + 20 = 440 J/K. Energy conservation gives 100(100 - T) = 440(T - 20). Thus 18800 = 540T and T = 34.81 degrees Celsius. The temperature lies between the two starting temperatures, as a basic sign check requires.
2. Recalculate while ignoring the container and explain the difference.
Read the worked answer
Ignoring the container gives 100(100 - T) = 420(T - 20), so T = 18400/520 = 35.38 degrees Celsius. This is about 0.57 degrees higher. The omitted container actually absorbs energy while warming from 20 degrees; ignoring that energy leaves too much predicted for the water. Its role is described by heat capacity in J/K, not by adding an arbitrary extra mass of water.