Ice at -20°C is added to 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it

Q: Why doesn't all the ice melt?

The heat provided by the water cooling down to $0^\circ\text{C}$ is only enough to warm all the ice to $0^\circ\text{C}$ and melt a portion of it. Once the water reaches $0^\circ\text{C}$, it can't provide any more heat to melt the rest of the ice.

Q: Why do we calculate the heat to warm the entire mass of ice, but only melt a part of it?

Heat transfer happens uniformly. Before any ice can melt, the entire chunk of ice must first reach its melting point ($0^\circ\text{C}$). Only after the whole mass is at $0^\circ\text{C}$ does the phase change (melting) begin.

Q: What if the heat lost by water wasn't even enough to bring the ice to $0^\circ\text{C}$?

In that case, the final temperature of the mixture would be below $0^\circ\text{C}$, and all the water would freeze into ice!

Q: Why is the specific heat of ice different from water?

Specific heat depends on the molecular structure and phase of the substance. Ice has a rigid crystalline structure, making it easier to raise its temperature compared to liquid water, which has more degrees of freedom.

Q: Can we use calories instead of Joules?

Yes, as long as you are consistent. If you use calories for specific heat and latent heat, your heat energy will be in calories. The final mass will remain the same.

Question

Ice at $-20^{\circ}	ext{C}$ is added to $50	ext{ g}$ of water at $40^{\circ}	ext{C}$. When the temperature of the mixture reaches $0^{\circ}	ext{C}$, it is found that $20	ext{ g}$ of ice is still unmelted. The amount of ice added to the water was close to
(Take, specific heat of water $= 4.2	ext{ J/g/}^{\circ}	ext{C}$, specific heat of ice $= 2.1	ext{ J/g/}^{\circ}	ext{C}$ and heat of fusion of water at $0^{\circ}	ext{C} = 334	ext{ J/g}$)

Options:

(A) $40	ext{ g}$

(B) $50	ext{ g}$

(C) $60	ext{ g}$

(D) $100	ext{ g}$

Sigma Percentile · Accepted Answer

### The Setup: A Battle of Temperatures

Imagine a classic thermal showdown. On one side, we have $50	ext{ g}$ of water sitting comfortably at $40^\circ	ext{C}$. On the other side, we drop in an unknown mass of ice, let's call it $x	ext{ g}$, chilling at $-20^\circ	ext{C}$.

When these two meet, a rapid exchange of heat begins. The water wants to cool down, and the ice wants to warm up and melt.

The problem gives us a crucial piece of intel: the final temperature of the mixture settles exactly at $0^\circ	ext{C}$, and there is still $20	ext{ g}$ of ice left unmelted. This detail is the key to unlocking the entire puzzle.

### The Master Equation: Principle of Calorimetry

In any isolated thermal system, energy is conserved. The heat lost by the hotter object must be exactly equal to the heat gained by the colder object.

$$Q_{	ext{lost}} = Q_{	ext{gained}}$$

Here, the water is the hot body losing heat, and the ice is the cold body gaining heat. Let's break down each side of this equation.

### Calculating the Heat Lost

The water starts at $40^\circ	ext{C}$ and cools down to $0^\circ	ext{C}$. The formula for heat transfer during a temperature change is:

$$Q = m s \Delta T$$

Plugging in the values for water:

$$Q_{	ext{lost}} = 50 	imes 4.2 	imes (40 - 0)$$

$$Q_{	ext{lost}} = 8400	ext{ J}$$

So, the water releases $8400	ext{ J}$ of heat energy into the system.

### Calculating the Heat Gained: The Tricky Part

Now, let's look at the ice. This is where many students make a **silly mistake**. The ice doesn't just melt instantly.

First, the entire mass of ice ($x	ext{ g}$) must warm up from $-20^\circ	ext{C}$ to $0^\circ	ext{C}$.

$$Q_{	ext{warming}} = x 	imes 2.1 	imes (0 - (-20)) = 42x$$

Once all the ice is at $0^\circ	ext{C}$, it starts to melt. But wait! The problem states that $20	ext{ g}$ of ice remains unmelted. This means only $(x - 20)	ext{ g}$ of ice actually underwent the phase change into water.

$$Q_{	ext{melting}} = (x - 20) 	imes 334$$

The total heat gained by the ice is the sum of these two processes:

$$Q_{	ext{gained}} = 42x + 334(x - 20)$$

$$Q_{	ext{gained}} = 42x + 334x - 6680$$

$$Q_{	ext{gained}} = 376x - 6680$$

### The Final Showdown: Equating and Solving

Now we bring it all together. We equate the heat lost by the water to the total heat gained by the ice.

$$8400 = 376x - 6680$$

Let's isolate $x$:

$$376x = 8400 + 6680$$

$$376x = 15080$$

$$x = \frac{15080}{376} \approx 40.10	ext{ g}$$

The calculated mass is approximately $40.10	ext{ g}$. Looking at our options, the closest value is **$40	ext{ g}$**.

This problem beautifully illustrates how heat transfer happens in stages. Always remember to account for warming up before phase changes!

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Visualized Solution

$T_{final} = 0^{\circ} C$

$Principle of Calorimetry$

$Heat Lost by Water$

$Calculating Q_{lost}$

$Heat Gained by Ice$

$Calculating Q_{gained}$

$Equating Heat$

$Final Answer$

$The Way Forward$

The Sigma Insight: Calorimetry

The Setup

The Master Equation

Calculating the Heat Lost

Calculating the Heat Gained

The Final Showdown

Similar Questions

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$M$ grams of steam at $10 0^{\circ} C$ is mixed with $200 g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $4 0^{\circ} C$ [heat of vaporisation of water is $540 cal/g$ and heat of fusion of ice is $80 cal/g$ ], the value of $M$ is ... .

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tfinal​=0∘C

.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 Water

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

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

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

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