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

Answer:

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$

Step-by-step 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:

$c = \frac{Q}{m\cdot \Delta T}$

$c = \frac{1600\,J}{(2\,kg)\cdot (13^{\circ}C-5^{\circ}C)}$

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$