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Animated Solution for Physics - Properties of Solids and Liquids: 300 g of water at $2 5^{\circ} C$ is added to 100 g of ice at $0^{\circ} C$ . The final temperature of the mixture is ......... $^{\circ} C$ .

Q: Why did we calculate the heat released by water cooling to exactly 0°C?

We do this to find the 'maximum available heat' without the water changing its phase. 0°C is the boundary where ice exists. If water cools to 0°C and still hasn't melted all the ice, the temperature cannot drop further without the water itself freezing, which won't happen because the ice is also at 0°C.

Q: What if the heat released by water was greater than 8000 cal?

If the water had enough heat to melt all the ice (e.g., if it was at 50°C), then all the ice would melt, and the remaining heat would raise the temperature of the entire 400g of water above 0°C. We would then use the equation $Q_{\text{remaining}} = (m_w + m_i) \times s_w \times (T_f - 0)$ to find the final temperature $T_f$.

Q: How much ice actually melted in this process?

The water provided 7500 cal of heat. Since melting 1 gram of ice takes 80 cal, the mass of ice melted is $m = \frac{7500}{80} = 93.75\text{ g}$.

Q: What is the final composition of the mixture?

The final mixture consists of the original 300 g of water plus the 93.75 g of melted ice, making 393.75 g of water. The remaining ice is $100 - 93.75 = 6.25\text{ g}$. Both are at 0°C.

Q: Why doesn't the temperature drop below 0°C?

For the temperature to drop below 0°C, the entire mixture would have to be ice, and it would need to lose heat to an even colder environment. Here, the coldest component is the ice at 0°C, so the mixture cannot get colder than that.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Mass of water, $m_{w} = 300 g$
Initial temperature of water, $T_{w} = 2 5^{\circ} C$
Mass of ice, $m_{i} = 100 g$
Initial temperature of ice, $T_{i} = 0^{\circ} C$

Principle of Calorimetry

Heat lost by hot body = Heat gained by cold body
Water will lose heat and cool down.
Ice will gain heat and melt.

Maximum Heat Released by Water

Let's calculate the maximum heat water can release if it cools down to $0^{\circ} C$ .
$Q_{released} = m_{w} s_{w} Δ T$

Calculating $Q_{released}$

$Q_{released} = 300 \times 1 \times (25 - 0)$
$Q_{released} = 7500 cal$

Heat Required to Melt All Ice

Now, let's find out how much heat is required to melt the entire $100 g$ of ice at $0^{\circ} C$ .
$Q_{required} = m_{i} L_{f}$

Calculating $Q_{required}$

$Q_{required} = 100 \times 80$
$Q_{required} = 8000 cal$

Comparing Heats

$Q_{released} = 7500 cal$
$Q_{required} = 8000 cal$
$Q_{released} < Q_{required}$

Final State and Temperature

Since all the ice does not melt, ice and water will coexist in the final mixture.
The temperature at which ice and water coexist in thermal equilibrium is $0^{\circ} C$ .

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The Sigma Insight: Calorimetry

Solution Diagram

The Battle of Heat

Water vs. Ice

Imagine a classic thermal showdown:

300 g

of relatively warm water at

2 5^{\circ} C

meets

100 g

of freezing ice at

0^{\circ} C

. What happens next is governed by the fundamental principle of calorimetry: Heat lost by the hot body equals heat gained by the cold body.

But before we jump into writing equations, we need to ask a crucial question: Does the water have enough thermal energy to melt all the ice?

Calculating the Thermal Arsenal

Let's first calculate the maximum amount of heat the water can give up. The water can cool down from

2 5^{\circ} C

to a minimum of

0^{\circ} C

before it starts freezing itself. The heat released in this process is given by:

Q_{released} = m_{w} s_{w} Δ T

Substituting the values (

m_{w} = 300 g

s_{w} = 1 cal/g^{\circ} C

Δ T = 2 5^{\circ} C

Q_{released} = 300 \times 1 \times 25 = 7500 cal

So, the water has a thermal arsenal of

7500 cal

The Ice's Demand

Now, let's see how much heat the ice demands to completely melt into water at

0^{\circ} C

. The heat required for this phase change is:

Q_{required} = m_{i} L_{f}

Substituting the values (

m_{i} = 100 g

L_{f} = 80 cal/g

Q_{required} = 100 \times 80 = 8000 cal

The Verdict

We have a situation! The water can only provide

7500 cal

, but the ice needs

8000 cal

to melt completely.

Q_{released} < Q_{required}

Because the available heat is less than the required heat, not all the ice will melt. The water will cool down to

0^{\circ} C

, giving all its

7500 cal

to the ice. This heat will melt a portion of the ice (specifically,

7500/80 = 93.75 g

of ice).

In the final state, we will have a mixture of water (the original

300 g

plus the newly melted

93.75 g

) and the remaining unmelted ice (

100 - 93.75 = 6.25 g

Whenever ice and water coexist in thermal equilibrium, the temperature of the system must be exactly at the melting point. Therefore, the final temperature of the mixture is $0^{\circ} C$ .

Common Traps & FAQs

Why did we calculate the heat released by water cooling to exactly 0°C?

We do this to find the 'maximum available heat' without the water changing its phase. 0°C is the boundary where ice exists. If water cools to 0°C and still hasn't melted all the ice, the temperature cannot drop further without the water itself freezing, which won't happen because the ice is also at 0°C.

What if the heat released by water was greater than 8000 cal?

If the water had enough heat to melt all the ice (e.g., if it was at 50°C), then all the ice would melt, and the remaining heat would raise the temperature of the entire 400g of water above 0°C. We would then use the equation

Q_{remaining} = (m_{w} + m_{i}) \times s_{w} \times (T_{f} - 0)

to find the final temperature

T_{f}

How much ice actually melted in this process?

The water provided 7500 cal of heat. Since melting 1 gram of ice takes 80 cal, the mass of ice melted is

m = \frac{7500}{80} = 93.75 g

What is the final composition of the mixture?

The final mixture consists of the original 300 g of water plus the 93.75 g of melted ice, making 393.75 g of water. The remaining ice is

100 - 93.75 = 6.25 g

. Both are at 0°C.

Why doesn't the temperature drop below 0°C?

For the temperature to drop below 0°C, the entire mixture would have to be ice, and it would need to lose heat to an even colder environment. Here, the coldest component is the ice at 0°C, so the mixture cannot get colder than that.

Animated Solution for Physics - Properties of Solids and Liquids: 300 g of water at $2 5^{\circ} C$ is added to 100 g of ice at $0^{\circ} C$ . The final temperature of the mixture is ......... $^{\circ} C$ .

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Principle of Calorimetry

Maximum Heat Released by Water

Calculating $Q_{released}$

Heat Required to Melt All Ice

Calculating $Q_{required}$

Comparing Heats

Final State and Temperature

The Sigma Insight: Calorimetry

The Battle of Heat

Calculating the Thermal Arsenal

The Ice's Demand

The Verdict

Similar Questions

When $M_{1}$ gram of ice at $- 1 0^{\circ} C$ (specific heat $= 0.5 \text{ cal g}^{-1} ^\circ\text{C}^{-1}$ ) is added to $M_{2}$ gram of water at $5 0^{\circ} C$ , finally no ice is left and the water is at $0^{\circ} C$ . The value of latent heat of ice, in $cal g^{- 1}$ is

$M$ 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 ... .

When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C will be

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Principle of Calorimetry

Maximum Heat Released by Water

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

Heat Required to Melt All Ice

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

Comparing Heats

Final State and Temperature

The Sigma Insight: Calorimetry

The Battle of Heat

Calculating the Thermal Arsenal

The Ice's Demand

The Verdict

Similar Questions

When 100 g of a liquid A at 100°C is added to 50 g of a liquid B at temperature 75°C, the temperature of the mixture becomes 90°C. The temperature of the mixture, if 100 g of liquid A at 100°C is added to 50 g of liquid B at 50°C will be

Calculating $Q_{released}$

Calculating $Q_{required}$