A certain amount of ice is supplied heat at a constant rate for 7 minutes. For the first one minute the temperature rises uniformly with time. Then, it remains constant for the next 4 minute and again the temperature rises at uniform rate for the last two minutes. Calculate the final temperature at the end of seven minutes.
(Given, L of ice `= 336 xx (10^3) J//kg` and specific heat of water `= 4200 J//kg-K`).

Explanation

To solve the problem, we need to analyze the heat transfer in three distinct phases over the 7 minutes.

### Step 1: Understand the phases of heating
1. **Phase 1 (0 to 1 minute)**: Ice is heated, and its temperature rises uniformly.
2. **Phase 2 (1 to 5 minutes)**: The temperature remains constant as the ice melts into water.
3. **Phase 3 (5 to 7 minutes)**: The temperature of the water rises uniformly.

### Step 2: Define the heat transfer in each phase
Let \( K \) be the rate of heat supplied (in Joules per second).

- **Phase 1**:
- Heat supplied, \( Q_1 = K \times 1 \) minute = \( K \times 60 \) seconds.

- **Phase 2**:
- Heat supplied, \( Q_2 = K \times 4 \) minutes = \( K \times 240 \) seconds.

- **Phase 3**:
- Heat supplied, \( Q_3 = K \times 2 \) minutes = \( K \times 120 \) seconds.

### Step 3: Analyze the heat transfer
1. **Phase 1**: Heating the ice from its initial temperature \( T_1 \) (assumed to be 0°C) to 0°C:
\[ Q_1 = m \cdot c_{ice} \cdot (0 - T_1) = m \cdot c_{ice} \cdot (0) \]
Here, \( c_{ice} \) is the specific heat of ice.

2. **Phase 2**: Melting the ice at 0°C:
\[ Q_2 = m \cdot L_f \]
where \( L_f = 336 \times 10^3 \, J/kg \) is the latent heat of fusion.

3. **Phase 3**: Heating the water from 0°C to the final temperature \( T_2 \):
\[ Q_3 = m \cdot c_{water} \cdot (T_2 - 0) = m \cdot 4200 \cdot T_2 \]
where \( c_{water} = 4200 \, J/kg \cdot K \) is the specific heat of water.

### Step 4: Set up the energy balance equations
From the heat supplied in each phase, we can set up the following equations:

1. From Phase 1 and Phase 2:
\[ K \times 60 = m \cdot c_{ice} \cdot (0 - T_1) + m \cdot L_f \]

2. From Phase 2 and Phase 3:
\[ K \times 240 = m \cdot L_f \]

3. From Phase 3:
\[ K \times 120 = m \cdot 4200 \cdot T_2 \]

### Step 5: Solve for \( T_2 \)
Using the second equation, we can express \( K \):
\[ K = \frac{m \cdot L_f}{240} \]
Substituting \( K \) into the third equation:
\[ \frac{m \cdot L_f}{240} \times 120 = m \cdot 4200 \cdot T_2 \]
This simplifies to:
\[ \frac{m \cdot L_f}{2} = m \cdot 4200 \cdot T_2 \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{L_f}{2} = 4200 \cdot T_2 \]
Now substituting \( L_f = 336 \times 10^3 \):
\[ \frac{336 \times 10^3}{2} = 4200 \cdot T_2 \]
\[ 168 \times 10^3 = 4200 \cdot T_2 \]
Solving for \( T_2 \):
\[ T_2 = \frac{168 \times 10^3}{4200} = 40 \, °C \]

### Final Answer
The final temperature at the end of seven minutes is **40°C**.