Question

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170gm of water at room temperature. Subsequently, the temperature of the system is found to be ${{75}^{\circ }}C$. T is given by:
(Given : room temperature = ${{30}^{\circ }}C$, specific heat of copper = 0.1cal/gm$^{\circ }C$)
$\text{A}\text{. }{{825}^{\circ }}C$
$\text{B}\text{. }{{800}^{\circ }}C$
$\text{C}\text{. }{{885}^{\circ }}C$
$\text{D}\text{. 1}{{250}^{\circ }}C$

Answer 1

Hint: In this case, the heat lost by the copper ball will be absorbed by the copper calorimeter and the water insides. Hence, the heat lost by the ball is equal to the sum of heat gained by the water and the calorimeter. To equate the heat energies, use the formula for heat lost or gained by a body when its temperature changes by $\Delta T$, i.e. $Q=mc\Delta T$. The specific heat of water is 1cal/gm$^{\circ }C$.
Formula used:
$Q=mc\Delta T$

Complete step-by-step solution:
It is given that a 100gm copper ball at temperature T is put into a 100gm calorimeter that is filled with 170gm of water. It is given that the calorimeter and the water are at room temperature.
Let us first analyze the given case. The temperature of the copper ball is higher than the calorimeter and water. When bodies with different temperatures come into contact, there is an exchange between the bodies (i.e. within the system) and the temperature of all the bodies in contact becomes the same. At this point, we say that the system of these bodies attained an equilibrium.
When two bodies are in contact, the body at a higher temperature loses heat, and the body at lower temperature gains heat. The body at lower temperature absorbs the heat given by the first body and increases its temperature. This way the two attain the same temperatures.
Heat lost or taken by a body for a change in its temperature $\Delta T$ is given as $Q=mc\Delta T$.
Here, m and ‘c’ are the mass and specific heat of the material of the body.
Therefore, when the copper ball comes in contact with the copper calorimeter and the water inside, it loses heat. This heat will be absorbed by the calorimeter and water and their temperature will increase. It is given that the equilibrium temperature of the three bodies is ${{75}^{\circ }}C$.
It is given that the specific heat of copper is 0.1cal/gm$^{\circ }C$. The specific heat of water is 1cal/gm$^{\circ }C$
Therefore, the heat lost by the copper ball is ${{Q}_{b}}={{m}_{b}}{{c}_{b}}(T-75)=100\times 0.1\times (T-75)$.
$\Rightarrow {{Q}_{b}}=10\times (T-75)$
The heat absorbed by the calorimeter is ${{Q}_{c}}={{m}_{c}}{{c}_{c}}(75-30)=100\times 0.1\times 45=450cal$.
The heat absorbed by the water is ${{Q}_{w}}={{m}_{w}}{{c}_{w}}(75-30)=170\times 1\times 45=7650cal$.
And we know that ${{Q}_{b}}={{Q}_{c}}+{{Q}_{w}}$
$\Rightarrow 10(T-75)=450+7650$
$\Rightarrow 10T-750=8100$
$\Rightarrow T=\dfrac{8850}{10}={{885}^{\circ }}C$.
Hence, the correct option is C.


Note: Always remember that when bodies, at least one at a different temperature, come into contact, the net heat change always is zero (if the surrounding absorbs no heat).
This means that the heat lost by the bodies at higher temperature is equal to the heat gained (absorbed) by the bodies at lower temperatures.
Therefore, in the given question we have neglected the heat absorbed by the outer surrounding.