Question

A 2 kg metal cylinder is supplied with 1600J of energy to heat it from 5°C to 13°C. What is the specific heat capacity of the metal?

Answer 1

`c = 100\,(J)/(kg\cdot ^(\circ)C)`

Explanation:

According to the First Law of Thermodynamics, the heat received by the metal cylinder is equal to the change in the internal energy. That is:

`Q = \Delta U`

`Q = m \cdot c \cdot \Delta T`

The specific heat is clear in the previous expression and finally computed:

`c = (Q)/(m\cdot \Delta T)`

`c = (1600\,J)/((2\,kg)\cdot (13^(\circ)C-5^(\circ)C))`

`c = 100\,(J)/(kg\cdot ^(\circ)C)`