A metal ball of mass 0.1 kg is heated upto 500°C and dropped into a vessel of heat capacity 800 JK^-

Q: Why do we add the heat capacity of the vessel to the water's heat gain?

Because the water is inside the vessel, they are in thermal contact. When the hot ball transfers heat, it warms up both the water and the vessel simultaneously. Ignoring the vessel would violate the conservation of energy.

Q: What is the difference between specific heat capacity and heat capacity?

Specific heat capacity ($s$) is the heat required to raise the temperature of 1 kg of a substance by 1°C. Heat capacity ($C$) is the heat required to raise the temperature of the *entire object* by 1°C. They are related by $C = ms$.

Q: Why is the percentage increment calculated with respect to $30^\circ\text{C}$ and not the final temperature?

Percentage change is always calculated relative to the *initial* value unless specified otherwise. Here, the initial state of the water was $30^\circ\text{C}$, so it forms the base of our percentage calculation.

Q: Can we use Kelvin instead of Celsius for the temperature differences?

Yes, absolutely! A temperature difference of $1^\circ\text{C}$ is exactly equal to a temperature difference of $1\text{ K}$. So, $\Delta T$ in Celsius is numerically identical to $\Delta T$ in Kelvin.

Q: Why did we approximate $21.3\%$ to $20\%$?

In many competitive exams, options are spaced out, and the question explicitly asks for the 'approximate' percentage increment. $21.3\%$ is significantly closer to $20\%$ than to $15\%$ or $25\%$, making it the safest logical choice.

Question

A metal ball of mass $0.1	ext{ kg}$ is heated upto $500^\circ	ext{C}$ and dropped into a vessel of heat capacity $800	ext{ JK}^{-1}$ and containing $0.5	ext{ kg}$ water. The initial temperature of water and vessel is $30^\circ	ext{C}$. What is the approximate percentage increment in the temperature of the water?
[Take, specific heat capacities of water and metal are respectively $4200	ext{ Jkg}^{-1}	ext{K}^{-1}$ and $400	ext{ Jkg}^{-1}	ext{K}^{-1}$]

Options:

(A) $25\%$

(B) $15\%$

(C) $30\%$

(D) $20\%$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a classic thermodynamics experiment. We have a scorching hot metal ball, weighing $0.1 	ext{ kg}$, heated to a blistering $500^\circ	ext{C}$. Below it waits a vessel containing $0.5 	ext{ kg}$ of water. Both the water and the vessel are resting at a comfortable room temperature of $30^\circ	ext{C}$.

When we drop the hot ball into the water, a rapid exchange of thermal energy begins. The ball will cool down, and the water along with the vessel will heat up until they all reach a common final temperature, let's call it $T$. Our ultimate goal is to find out by what percentage the water's temperature increases.

### The Master Equation: Principle of Calorimetry

To solve this, we rely on the fundamental **Principle of Calorimetry**, which is essentially the law of conservation of energy applied to heat transfer. Assuming no heat is lost to the surrounding environment, we can state:

$$ 	ext{Heat Lost by Hot Body} = 	ext{Heat Gained by Cold Bodies} $$

Let's break this down into two parts. First, the heat lost by the metal ball. The formula for heat transfer when there is a temperature change is $\Delta Q = ms\Delta T$.

For the metal ball, the mass $m_b = 0.1 	ext{ kg}$, the specific heat $s_b = 400 	ext{ Jkg}^{-1}	ext{K}^{-1}$, and the change in temperature is $(500 - T)$.

$$ Q_{	ext{lost}} = 0.1 	imes 400 	imes (500 - T) = 40(500 - T) $$

### Heat Gained by the System

Now, who is absorbing all this thermal energy? It's not just the water; the vessel holding the water also heats up. We must account for both.

For the water, we use the same formula $\Delta Q = ms\Delta T$. The mass of water $m_w = 0.5 	ext{ kg}$, its specific heat $s_w = 4200 	ext{ Jkg}^{-1}	ext{K}^{-1}$, and its temperature change is $(T - 30)$.

For the vessel, we are directly given its **heat capacity** $C_v = 800 	ext{ JK}^{-1}$. Notice that heat capacity is already the product of mass and specific heat ($C = ms$). So, the heat gained by the vessel is simply $C_v \Delta T$.

Total heat gained by the water and vessel is:

$$ Q_{	ext{gained}} = 0.5 	imes 4200 	imes (T - 30) + 800 	imes (T - 30) $$

$$ Q_{	ext{gained}} = 2100(T - 30) + 800(T - 30) = 2900(T - 30) $$

### Final Calculation

Now, we bring it all together by equating the heat lost to the heat gained:

$$ 40(500 - T) = 2900(T - 30) $$

We can simplify this by dividing both sides by 10:

$$ 4(500 - T) = 290(T - 30) $$

Expanding the brackets:

$$ 2000 - 4T = 290T - 8700 $$

Rearranging to solve for $T$:

$$ 294T = 10700 $$

$$ T = \frac{10700}{294} \approx 36.39^\circ	ext{C} $$

The final equilibrium temperature of the system is approximately $36.39^\circ	ext{C}$.

The question asks for the **approximate percentage increment** in the temperature of the water. The initial temperature was $30^\circ	ext{C}$, and the increment is $\Delta T = 36.39 - 30 = 6.39^\circ	ext{C}$.

$$ \% 	ext{ Increment} = \left( \frac{6.39}{30} ight) 	imes 100 \approx 21.3\% $$

Looking at our options, $21.3\%$ is closest to **$20\%$**. Therefore, the correct option is (d).

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Principle of Calorimetry$

$Heat Lost by the Metal Ball$

$Heat Gained by Water and Vessel$

$Equating Heat Lost and Gained$

$Solving for Final Temperature T$

$Calculating Percentage Increment$

$The Way Forward$

The Sigma Insight: Calorimetry

Analyzing the Setup

The Master Equation

Heat Gained by the System

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Principle of Calorimetry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Lost by the Metal Ball

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Gained by Water and Vessel

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Heat Lost and Gained

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Final Temperature T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Percentage Increment

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward