Question

A solid of mass ${\rm{50g}}$ at $150^\circ {\rm{C}}$ is placed in $100g$ of water at $11^\circ {\rm{C}}$ when the final temperature recorded is $20^\circ {\rm{C}}$. Find the specific heat capacity of the solid. (Specific heat capacity of water $4.2\;{\rm{J/g}}^\circ {\rm{C}}$ )

Answer 1

Hint: According to the principle of calorimetry, heat lost by one body is equal to the heat gained by another body when both bodies are both solid and other is fluid. The thermal heat capacity of the body plays an important role in the gain or loss of heat.

Complete step-by-step answer:
Given: The mass of the solid is $M = 50\;{\rm{g}}$
The temperature of the solid is $T = 150^\circ {\rm{C}}$.
The mass of the water is $m = 100\;{\rm{g}}$.
The temperature of the water is $t = 11^\circ {\rm{C}}$.
The final temperature of the mixture is${t_1} = 20^\circ {\rm{C}}$.
According to the principle of calorimetry:
Heat gained by the water=Heat lost by the solid
\[\begin{array}{l}
MC({t_1} - t) = mc(T - {t_1})\\
100\;{\rm{g}} \times 4.2\dfrac{{{\rm{KJ}}}}{{{\rm{kg}}\,{\rm{K}}}} \times (20 - 11)^\circ {\rm{C}} = 50\,{\rm{g}} \times c \times (150 - 20)^\circ {\rm{C}}\\
c = 0.58\dfrac{{{\rm{KJ}}}}{{{\rm{kg}}\,{\rm{K}}}}
\end{array}\]
Here C is the specific heat capacity of water and c is the specific heat capacity of solid.

Therefore, the specific heat capacity of the solid is \[0.58\dfrac{{{\rm{KJ}}}}{{{\rm{kg}}\,{\rm{K}}}}\].

Note: If the specific heat capacity of a body is high the time required to change the temperature of the body is very high. For example, in the water, the time required to cool the water from high temperature to room temperature is very large.
This question can also be solved by taking both the bodies as system and insulated from the surrounding and hence we can apply the first law of thermodynamics and can equate the internal energies of the body and taking the heat entered to the surrounding or lost to the surrounding is zero.