Question

Temperature of a metal ball is $30^{\circ} C$. When an energy of $3000$J is supplied its temperature rises by $40^{\circ} C$. Calculate its heat capacity.

Answer 1

Hint: Heat capacity is the proportion of heat energy transported to an object to enhance its temperature. The ability of a substance to receive heat energy; the amount of heat needed to enhance the temperature of one mole of a matter by one degree Celsius without any modification of phase.


Complete step-by-step solution:
Given: Initial temperature, $T_{i} = 30^{\circ} C$
Final temperature, $T_{f} = 40^{\circ} C$
Heat required, $Q = 3000 J$
Whenever there is change in temperature, heat is required.
$Q = ms (dt)$
$\implies Q = ms (T_{f} – T_{i})$
m is the mass.
s is the specific heat capacity.
We need to calculate the heat capacity.
ms is the heat capacity.
The formula for heat capacity is:
$ms = \dfrac{Q}{ (T_{f} – T_{i}) }$
Put all values in the above formula:
$ms = \dfrac{3000}{ (40 – 30)^{\circ} C }$
$\implies ms = \dfrac{3000}{ 10^{\circ} C } \times \dfrac{1 cal}{ 4.182 J}$
$\implies ms = 71.7 cal^{\circ}C$
So, the heat capacity is $71.7 cal^{\circ}C$.

Note:Various substances have various specific heat capacities. The substance that has a higher specific heat capacity varies its temperature gradually. The substance that has a small specific heat capacity varies its temperature fast.