Aluminium container of mass of `10` g contains `200` g of ice at`-20^(@)C`. Heat is added to the system at the rate of `100` calories per second. What is the temperature of the system after four minutes? Draw a rough sketch showing the variation of the temperature of the system as a function of time . Given:
Specific heat of ice = `0.5"cal"g^(-1)(.^(@)C)^(-1)`
Specific heat of aluminium = `0.2"cal"g^(-1)(.^(@)C)^(-1)`
Latent heat of fusion of ice = `80"cal"g^(-1)`

Explanation

To solve the problem step by step, we will calculate the total heat supplied to the system, and then determine how this heat is distributed among the different components (the aluminum container and the ice) to find the final temperature of the system.

### Step 1: Calculate the total heat supplied in 4 minutes
The rate of heat supplied is given as \(100\) calories per second.

\[ \text{Total time} = 4 \text{ minutes} = 4 \times 60 = 240 \text{ seconds} \]

\[ \text{Total heat supplied} = \text{Rate} \times \text{Time} = 100 \, \text{cal/s} \times 240 \, \text{s} = 24000 \, \text{calories} \]

### Step 2: Set up the heat balance equation
Let \(T\) be the final temperature of the system. The heat absorbed by the system can be expressed as the sum of the heat absorbed by the aluminum container, the heat required to raise the temperature of the ice to \(0^\circ C\), the latent heat required to melt the ice, and the heat required to raise the temperature of the resulting water to \(T\).

The heat absorbed by the aluminum container:
\[ Q_{\text{Al}} = m_{\text{Al}} \cdot s_{\text{Al}} \cdot (T - T_{\text{initial, Al}}) \]
Where:
- \(m_{\text{Al}} = 10 \, \text{g}\)
- \(s_{\text{Al}} = 0.2 \, \text{cal/g}^\circ C\)
- \(T_{\text{initial, Al}} = T_{\text{initial, ice}} = -20^\circ C\)

\[ Q_{\text{Al}} = 10 \cdot 0.2 \cdot (T - (-20)) = 2(T + 20) \]

The heat required to raise the temperature of the ice from \(-20^\circ C\) to \(0^\circ C\):
\[ Q_{\text{ice}} = m_{\text{ice}} \cdot s_{\text{ice}} \cdot (0 - (-20)) \]
Where:
- \(m_{\text{ice}} = 200 \, \text{g}\)
- \(s_{\text{ice}} = 0.5 \, \text{cal/g}^\circ C\)

\[ Q_{\text{ice}} = 200 \cdot 0.5 \cdot 20 = 2000 \, \text{calories} \]

The latent heat required to melt the ice:
\[ Q_{\text{latent}} = m_{\text{ice}} \cdot L_f \]
Where:
- \(L_f = 80 \, \text{cal/g}\)

\[ Q_{\text{latent}} = 200 \cdot 80 = 16000 \, \text{calories} \]

The heat required to raise the temperature of the water (after melting) from \(0^\circ C\) to \(T\):
\[ Q_{\text{water}} = m_{\text{water}} \cdot s_{\text{water}} \cdot (T - 0) \]
Where:
- \(m_{\text{water}} = 200 \, \text{g}\)
- \(s_{\text{water}} = 1 \, \text{cal/g}^\circ C\)

\[ Q_{\text{water}} = 200 \cdot 1 \cdot T = 200T \]

### Step 3: Combine all heat absorbed and set equal to heat supplied
Total heat absorbed:
\[ Q_{\text{total}} = Q_{\text{Al}} + Q_{\text{ice}} + Q_{\text{latent}} + Q_{\text{water}} \]
\[ Q_{\text{total}} = 2(T + 20) + 2000 + 16000 + 200T \]

Setting this equal to the total heat supplied:
\[ 24000 = 2(T + 20) + 2000 + 16000 + 200T \]

### Step 4: Simplify and solve for \(T\)
\[ 24000 = 2T + 40 + 2000 + 16000 + 200T \]
\[ 24000 = 202T + 18040 \]
\[ 24000 - 18040 = 202T \]
\[ 5960 = 202T \]
\[ T = \frac{5960}{202} \approx 29.5^\circ C \]

### Final Answer
The final temperature of the system after four minutes is approximately \(29.5^\circ C\).

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