An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimete

Q: Why do we add the heat gained by the calorimeter?

The calorimeter is in direct physical contact with the water. As the water heats up, it transfers heat to the brass walls until they are at the same temperature. If we ignore the calorimeter, we violate the conservation of energy!

Q: How do we convert specific heat from J/kg-K to J/g-K?

Since 1 kg = 1000 g, a value of 394 J/kg-K means it takes 394 Joules to heat 1000 grams by 1 Kelvin. To find the heat for just 1 gram, you divide by 1000, giving 0.394 J/g-K.

Q: Why is the final temperature the same for all three components?

This is the definition of thermal equilibrium. Heat will continue to flow between objects in contact until they all reach the exact same temperature, at which point the net heat transfer becomes zero.

Q: Why didn't we convert Celsius to Kelvin for the temperature?

The formula uses $\Delta T$, which is the *change* in temperature. A change of $1^\circ\text{C}$ is exactly equal to a change of $1\text{ K}$. So, $100^\circ\text{C} - 21.5^\circ\text{C} = 78.5^\circ\text{C}$, which is exactly a difference of $78.5\text{ K}$.

Q: Where did the specific heat of water (4.18) come from?

The specific heat of liquid water is a fundamental physical constant ($4.18 \text{ J g}^{-1}\text{K}^{-1}$ or $1 \text{ cal g}^{-1}^\circ\text{C}^{-1}$). In JEE, you are often expected to have this standard value memorized if it isn't explicitly provided in the question.

Question

An unknown metal of mass $192	ext{ g}$ heated to a temperature of $100^\circ	ext{C}$ was immersed into a brass calorimeter of mass $128	ext{ g}$ containing $240	ext{ g}$ of water at a temperature of $8.4^\circ	ext{C}$. Calculate the specific heat of the unknown metal, if water temperature stabilises at $21.5^\circ	ext{C}$. (Take, specific heat of brass is $394	ext{ J kg}^{-1}	ext{K}^{-1}$)

Options:

(A) $916	ext{ J kg}^{-1}	ext{K}^{-1}$

(B) $654	ext{ J kg}^{-1}	ext{K}^{-1}$

(C) $1232	ext{ J kg}^{-1}	ext{K}^{-1}$

(D) $458	ext{ J kg}^{-1}	ext{K}^{-1}$

Sigma Percentile · Accepted Answer

Imagine you are standing in a laboratory, holding a scorching hot piece of unknown metal. It's at a blistering $100^\circ	ext{C}$. In front of you is a brass calorimeter, a specialized container designed to isolate heat, filled with cool water at $8.4^\circ	ext{C}$.

When you drop that hot metal into the water, a microscopic battle of kinetic energy begins. The fast-moving atoms of the metal collide with the slower-moving molecules of the water and the brass walls of the calorimeter. Heat, which is simply energy in transit, begins to flow.

### The Master Principle: Calorimetry

The universe loves balance. The fundamental law governing this thermal dance is the **Principle of Calorimetry**. It states a beautifully simple truth: assuming no heat escapes into the surrounding room, the total heat lost by the hot object must exactly equal the total heat gained by the cold objects.

Mathematically, we express the heat transferred as:
$$ Q = ms\Delta T $$
Where $m$ is the mass, $s$ is the specific heat capacity (how stubborn a material is to change its temperature), and $\Delta T$ is the change in temperature.

### Breaking Down the Heat Exchange

Let's look at the players in our system. First, the hot metal. It starts at $100^\circ	ext{C}$ and cools down to the final equilibrium temperature of $21.5^\circ	ext{C}$. 
The heat it loses is:
$$ Q_{	ext{lost}} = m_m s_m \Delta T_m $$
$$ Q_{	ext{lost}} = 192 	imes s_m 	imes (100 - 21.5) = 15072 	imes s_m $$

Now, who is absorbing this heat? It's a tag team: the water and the brass calorimeter itself! 
For the water, we know its specific heat is a standard $4.18 	ext{ J g}^{-1}	ext{K}^{-1}$. It warms up from $8.4^\circ	ext{C}$ to $21.5^\circ	ext{C}$.
$$ Q_{	ext{water}} = 240 	imes 4.18 	imes (21.5 - 8.4) = 13142 	ext{ J} $$

But here is where many students make a **silly mistake**—they forget the container! The brass calorimeter also warms up. We are given its specific heat as $394 	ext{ J kg}^{-1}	ext{K}^{-1}$. Notice the units! We must convert this to grams to match the rest of our data, making it $0.394 	ext{ J g}^{-1}	ext{K}^{-1}$.
$$ Q_{	ext{brass}} = 128 	imes 0.394 	imes (21.5 - 8.4) = 660.65 	ext{ J} $$

### The Grand Equation

Now, we bring it all together. The heat lost by the metal was entirely absorbed by the water and the brass.
$$ Q_{	ext{lost}} = Q_{	ext{water}} + Q_{	ext{brass}} $$
$$ 15072 	imes s_m = 13142 + 660.65 $$
$$ 15072 	imes s_m = 13802.65 $$

### The Final Calculation

All that remains is to isolate our unknown variable, $s_m$. 
$$ s_m = \frac{13802.65}{15072} \approx 0.916 	ext{ J g}^{-1}	ext{K}^{-1} $$

Wait, look at the options! They are all in $	ext{J kg}^{-1}	ext{K}^{-1}$. To convert our answer back to kilograms, we simply multiply by $1000$.
$$ s_m = 916 	ext{ J kg}^{-1}	ext{K}^{-1} $$

And there we have it! By carefully tracking the flow of energy and respecting the units, we've unveiled the thermal identity of our unknown metal.

Select Answer:

Visualized Solution

$Visualizing the Calorimeter Setup$

$Principle of Calorimetry$

$Heat Lost by the Metal$

$Heat Gained by Water$

$Heat Gained by Brass Calorimeter$

$Equating Heat Lost and Gained$

$Calculating Specific Heat$

$Final Answer$

$The Way Forward$

The Sigma Insight: Calorimetry

The Master Principle

Breaking Down the Heat Exchange

The Grand Equation

The Final Calculation

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Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is

Select Answer:

Visualized Solution

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

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

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

.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; } Calculating Specific Heat

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

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