A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100

Q: Why do we assume the calorimeter and water have the same initial temperature?

Since the water is filled in the calorimeter and left at room temperature, they are in thermal equilibrium with each other before the hot ball is dropped in.

Q: Why isn't the specific heat of water given in the problem?

The specific heat of water ($1 \text{ cal/g}^{\circ}\text{C}$ or $4200 \text{ J/kg}^{\circ}\text{C}$) is a standard physical constant that students are expected to memorize for JEE.

Q: What does 'water equivalent' mean, and do we need it here?

Water equivalent is the mass of water that would absorb the same amount of heat as the calorimeter for the same temperature rise. Here, we directly used the mass and specific heat of the calorimeter, so we didn't need to calculate its water equivalent separately.

Q: Why is the temperature change for the ball written as $(T - 75)$ instead of $(75 - T)$?

When equating Heat Lost to Heat Gained, we want both quantities to be positive. Since the ball cools down, its initial temperature $T$ is greater than the final temperature $75^{\circ}\text{C}$, making $(T - 75)$ a positive value.

Q: Could some heat be lost to the surroundings?

In reality, yes. However, in standard calorimetry problems unless specified otherwise, we assume the calorimeter is perfectly insulated and no heat is lost to the surroundings.

Question

A copper ball of mass $100 	ext{ g}$ is at a temperature $T$. It is dropped in a copper calorimeter of mass $100 	ext{ g}$, filled with $170 	ext{ g}$ of water at room temperature. Subsequently, the temperature of the system is found to be $75^{\circ}	ext{C}$. $T$ is (Given, room temperature $= 30^{\circ}	ext{C}$, specific heat of copper $= 0.1 	ext{ cal/g}^{\circ}	ext{C}$)

Options:

(A) $885^{\circ}	ext{C}$

(B) $1250^{\circ}	ext{C}$

(C) $825^{\circ}	ext{C}$

(D) $800^{\circ}	ext{C}$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Temperatures

Imagine you are in a laboratory, holding a scorching hot copper ball. Its mass is $100 	ext{ g}$, and its temperature is a mysterious $T$. You are about to drop this fiery sphere into a copper calorimeter.

The calorimeter itself has a mass of $100 	ext{ g}$ and is filled with $170 	ext{ g}$ of water. Both the water and the calorimeter are resting peacefully at a room temperature of $30^{\circ}	ext{C}$.

When you drop the ball in, a rapid exchange of thermal energy begins. The hot ball cools down, and the water and calorimeter heat up until they all reach a final equilibrium temperature of $75^{\circ}	ext{C}$. Our mission is to find that initial temperature $T$.

### The Master Equation: Principle of Calorimetry

To solve this, we rely on the fundamental principle of calorimetry: **Heat Lost = Heat Gained**.

Assuming our calorimeter is perfectly insulated, no heat escapes into the surroundings. All the thermal energy lost by the hot copper ball is entirely absorbed by the colder water and the copper calorimeter.

The formula for heat transfer is $Q = ms\Delta T$, where $m$ is the mass, $s$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

### Crunching the Numbers: Finding the Unknown

Let's set up our equation. For the copper ball, the heat lost is $m_B s_B (T - 75)$. For the water and the calorimeter, the heat gained is $m_w s_w (75 - 30) + m_c s_c (75 - 30)$.

Equating the two, we get:

$$100 	imes 0.1 	imes (T - 75) = 170 	imes 1 	imes (75 - 30) + 100 	imes 0.1 	imes (75 - 30)$$

Notice that we used $1 	ext{ cal/g}^{\circ}	ext{C}$ for the specific heat of water, a standard constant you should always remember!

Now, let's simplify the right side. The temperature change for both the water and the calorimeter is $45^{\circ}	ext{C}$.

$$10(T - 75) = 170 	imes 45 + 10 	imes 45$$

$$10(T - 75) = 7650 + 450$$

$$10(T - 75) = 8100$$

Dividing both sides by $10$, we find:

$$T - 75 = 810$$

Finally, adding $75$ to $810$, we arrive at our answer:

$$T = 885^{\circ}	ext{C}$$

The copper ball was initially at a blistering **$885^{\circ}	ext{C}$**!

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e S y s t e m$

$P r in c i pl eo f C a l or im e t r y$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g H e a tG ain e d$

$S im pl i f y in g t h e E q u a t i o n$

$F ina l T e m p er a t u r e$

$R e a l W or l d C o n s i d er a t i o n s$

The Sigma Insight: Calorimetry

The Setup

The Master Equation

Crunching the Numbers

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Visualized Solution

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