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 $75^{\circ} C$. $T$ is given by : (Given : room temperature = $30^{\circ}C$, specific heat of copper = $0.1\, cal\,g^{-1}\,^{\circ} C^{-1}$)

• A. $800^{\circ}C$
• B. $885^{\circ}C$
• C. $1250^{\circ}C$
• D. $825^{\circ}C$

Answer 1

The Correct Option is B

To solve this problem, we need to apply the principle of calorimetry, which states that the heat lost by the hot body is equal to the heat gained by the cooler bodies when there is no loss of heat to the surroundings.

Let's breakdown the problem:

1. The copper ball (hot body) of mass $m_1 = 100 \, \text{gm}$ is initially at an unknown temperature $T$.
2. A copper calorimeter of mass $m_2 = 100 \, \text{gm}$ is filled with $m_3 = 170 \, \text{gm}$ of water at room temperature of $30^{\circ} C$.
3. The final temperature of the mixture is $75^{\circ} C$.
4. Specific heat of copper $c_\text{copper} = 0.1 \, \text{cal g}^{-1}\,^{\circ} \text{C}^{-1}$ and specific heat of water is approximately $1 \, \text{cal g}^{-1} \,^{\circ} \text{C}^{-1}$.

Applying the formula:

\text{Heat lost by copper ball} = \text{Heat gained by calorimeter} + \text{Heat gained by water}

Mathematically,

m_1 c_\text{copper} (T - 75) = m_2 c_\text{copper} (75 - 30) + m_3 c_\text{water} (75 - 30)

Let's substitute the given values:

100 \times 0.1 \times (T - 75) = 100 \times 0.1 \times 45 + 170 \times 1 \times 45

This simplifies to:

10(T - 75) = 450 + 7650

10(T - 75) = 8100

Dividing both sides by 10:

T - 75 = 810

Adding 75 to both sides gives:

T = 885^{\circ} C

Therefore, the initial temperature $T$ of the copper ball is $885^{\circ} C$.