Question

Calculate the heat capacity of two cast-iron frying pans, one large and one small. The temperature of each pan is increased by 50 degrees. That requires an input of 18,150 J of energy for the small pan, and 90,700 J for the large pan

Answer 1

Heat capacity $c$ is a ratio of heat $Q$ absorbed by a material to the temperature change $\Delta T$. The heat absorbed by the large pan is 90,700 J and the temperature change is 50 degrees. As it is a difference temperature, the value is the same in kelvin and in celsius. Therefore, the heat capacity of the large pan is:

$c (\text{large pan)} = \frac{Q}{\Delta T} = \frac{90700}{50} = 1814$ J/°C.

The heat absorbed by the small pan is 18,150 J and the temperature change is again 50 degrees. Analogically, the heat capacity of the small pan is:

$c (\text{small pan)} = \frac{Q}{\Delta T} = \frac{18150}{50} = 363$ J/°C.