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 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be $75°\mathrm{C}$. T is given by :(Given: room temperature = $30°\mathrm{C}$, the specific heat of copper = $0.1\mathrm{cal}/\mathrm{gm}°\mathrm{C}$)

• A. $1250°\mathrm{C}$
• B. $825°\mathrm{C}$
• C. $800°\mathrm{C}$
• D. $885°\mathrm{C}$

Answer 1

The correct option is D

$885°\mathrm{C}$

The explanation for the correct answer:-

Option (D) $885°\mathrm{C}$:

Step 1: Given data

Mass of copper ball= $100\mathrm{gm}$

Room temperature= $30°\mathrm{C}$

Mass of water= $170\mathrm{gm}$

Specific heat of copper = $0.1\mathrm{cal}\mathrm{gm}°{\mathrm{C}}^{-1}$

Temperature of system= $75°\mathrm{C}$

Step 2: Heat absorbed by the bodies

We know that heat lost or taken by a body for a change in its temperature $\mathrm{\Delta }T$ is given as $Q=mc\mathrm{\Delta }T$.

Where m= mass of the body and c= specific heat of the body

As a result, when the copper ball comes in contact with the copper calorimeter and the water which is inside, it loses heat. This heat will now be absorbed by the calorimeter and the water and its temperature will increase. It is also given that the equilibrium temperature of the three bodies is $75°\mathrm{C}$, and we know that the specific heat of water is $1\mathrm{cal}\mathrm{gm}°{\mathrm{C}}^{-1}$

$\therefore$heat lost by the copper ball is

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{b}}={\mathrm{m}}_{\mathrm{b}}{\mathrm{c}}_{\mathrm{b}}(\mathrm{T}-75°\mathrm{C}) \\ \Rightarrow {\mathrm{Q}}_{\mathrm{b}}=100\times 0.1\times (\mathrm{T}-75) \\ \Rightarrow \mathrm{Qb}=10(\mathrm{T}-75) \end{gathered}$$

$\therefore$heat lost by the calorimeter is

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{c}}={\mathrm{m}}_{\mathrm{c}}{\mathrm{c}}_{\mathrm{c}}(75-30) \\ \Rightarrow {\mathrm{Q}}_{\mathrm{c}}=100\times 0.1\times 45 \\ \Rightarrow {\mathrm{Q}}_{\mathrm{c}}=450\mathrm{cal} \end{gathered}$$

.$\therefore$heat lost by the calorimeter is

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{w}}={\mathrm{m}}_{\mathrm{w}}{\mathrm{c}}_{\mathrm{w}}(75-30) \\ \Rightarrow {\mathrm{Q}}_{\mathrm{w}}=170\times 1\times 45 \\ \Rightarrow {\mathrm{Q}}_{\mathrm{w}}=7650\mathrm{cal} \end{gathered}$$

Step 3: Calculating the temperature

We know that, ${\mathrm{Q}}_{\mathrm{b}}={\mathrm{Q}}_{\mathrm{c}}+{\mathrm{Q}}_{\mathrm{w}}$

$$\begin{gathered} \therefore 10(\mathrm{T}-75)=450+7650 \\ \Rightarrow 10\mathrm{T}-750=8100 \\ \Rightarrow \mathrm{T}=885010 \\ \Rightarrow \mathrm{T}=885°\mathrm{C} \end{gathered}$$

Thus the temperature is $\mathrm{T}=885°\mathrm{C}$

Therefore, T is given by $\mathrm{T}=885°\mathrm{C}$, option (D) is the correct answer.