Question

In an experiment on the specific heat of a metal, a $ 0.20{\text{ }}kg $ block of the metal at $ 150{\text{ }}^\circ C $ is dropped in a copper calorimeter (of water equivalent $ 0.025{\text{ }}kg $ ) containing $ 150{\text{ }}c{m^3} $ of water at $ 27{\text{ }}^\circ C $ . The final temperature is $ 40{\text{ }}^\circ C $ . Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for the specific heat of the metal?

Answer 1

Hint: When the metal is dropped in the copper calorimeter containing water, its own temperature will drop. The temperature of the container and the water will increase and the equilibrium temperature can be calculated by balancing the heat transfer in the transfer.

Formula used
In this solution, we will use the following formula:
 $ Q = ms\Delta T $ where $ Q $ is the heat transferred to a substance of mass $ m $ , specific heat capacity $ s $ , and $ \Delta T $ is the temperature difference.

Complete step by step answer:
When the metal block is dropped in the container, it will lose its temperature since it will transfer heat to the water and the container. In contrast, the temperature of the water inside the container will increase and the temperature of the container will also increase.
We know that the final equilibrium temperature is $ 40{\text{ }}^\circ C $ . Since there are no heat energy losses, the heat transferred by the metal block $ {Q_m} $ will be completely gained by the water $ {Q_w} $ and the container $ {Q_c} $ . So, we can write
 $ {Q_m} = {Q_w} + {Q_c} $
Before using this formula, we need the mass of water in the calorimeter. Since the volume of water is $ 150{\text{ }}c{m^3} = 150 \times {10^{ - 6}}\,{m^3} $ , its mass can be calculated from the density of water $ (\rho = {10^3}\,{m^3}) $ as
 $ m = \rho \times V $
 $ \Rightarrow m = {10^3}\, \times 150 \times {10^{ - 6}} $
Which gives us the mass as:
 $ m = 150 \times {10^{ - 3}}\,kg $
Now the water equivalent of the calorimeter is $ 0.025{\text{ }}kg $ . Let $ {s_w} $ be the specific heat capacity of water. So, we can write in the heat balance equation:
 $ 0.20 \times s \times (40 - 150) = \left( {150 \times {{10}^{ - 3}} \times (4.2 \times {{10}^3}) \times (40 - 27)} \right) + \left( {0.025 \times (4.2 \times {{10}^3}) \times (40 - 27)} \right) $
Solving the above equation for $ s $ , we get
 $ s = 434\,J/kg.K $
If heat losses to the surrounding area are not negligible then the value of specific heat of metal will be less than the actual value since the metal will have a lower capacity to increase its temperature for a given amount of heat.

Note:
Since we know the water equivalent mass of the container, we can assume it to be the equivalent of an amount of water that has mass $ 0.025{\text{ }}kg $ . Also, after using the water equivalent mass, it will have the same specific heat capacity too which we need to use in the heat balance equation since we haven’t been given the specific heat capacity of the container.