In the above problem if heat is supplied at a constant rate of `q= 10 cal//min`, then plot temperature versus time graph.

Explanation

To solve the problem of plotting the temperature versus time graph when heat is supplied at a constant rate of \( q = 10 \, \text{cal/min} \), we will follow these steps:

### Step 1: Identify the phases of heating and phase changes
The heating process can be divided into several phases based on the state of the substance (ice, water, steam) and the temperature ranges:
1. Heating ice from \(-3^\circ C\) to \(0^\circ C\) (Phase A to B)
2. Melting ice at \(0^\circ C\) (Phase B to C)
3. Heating water from \(0^\circ C\) to \(100^\circ C\) (Phase C to D)
4. Boiling water at \(100^\circ C\) (Phase D to E)

### Step 2: Calculate the heat required for each phase
1. **Phase A to B**: Heating ice from \(-3^\circ C\) to \(0^\circ C\)
- Heat required: \( Q_{AB} = m \cdot c_{ice} \cdot \Delta T = m \cdot 0.5 \cdot 3 = 1.5m \, \text{cal} \)
- Given \( m = 42 \, \text{g} \), \( Q_{AB} = 1.5 \cdot 42 = 63 \, \text{cal} \)

2. **Phase B to C**: Melting ice at \(0^\circ C\)
- Heat required: \( Q_{BC} = m \cdot L_f = 42 \cdot 80 = 3360 \, \text{cal} \)

3. **Phase C to D**: Heating water from \(0^\circ C\) to \(100^\circ C\)
- Heat required: \( Q_{CD} = m \cdot c_{water} \cdot \Delta T = 42 \cdot 1 \cdot 100 = 4200 \, \text{cal} \)

4. **Phase D to E**: Boiling water at \(100^\circ C\)
- Heat required: \( Q_{DE} = m \cdot L_v = 42 \cdot 540 = 22680 \, \text{cal} \)

### Step 3: Calculate time for each phase
Using the formula \( T = \frac{Q}{q} \) where \( q = 10 \, \text{cal/min} \):

1. **Phase A to B**:
\[ T_{AB} = \frac{63}{10} = 6.3 \, \text{min} \]

2. **Phase B to C**:
\[ T_{BC} = \frac{3360}{10} = 336 \, \text{min} \]

3. **Phase C to D**:
\[ T_{CD} = \frac{4200}{10} = 420 \, \text{min} \]

4. **Phase D to E**:
\[ T_{DE} = \frac{22680}{10} = 2268 \, \text{min} \]

### Step 4: Calculate total time at each phase
- Total time at point B: \( T_B = T_{AB} = 6.3 \, \text{min} \)
- Total time at point C: \( T_C = T_{AB} + T_{BC} = 6.3 + 336 = 342.3 \, \text{min} \)
- Total time at point D: \( T_D = T_{AB} + T_{BC} + T_{CD} = 342.3 + 420 = 762.3 \, \text{min} \)
- Total time at point E: \( T_E = T_{AB} + T_{BC} + T_{CD} + T_{DE} = 762.3 + 2268 = 3030.3 \, \text{min} \)

### Step 5: Plot the temperature versus time graph
1. **From A to B**: The temperature increases linearly from \(-3^\circ C\) to \(0^\circ C\) over \(6.3 \, \text{min}\).
2. **From B to C**: The temperature remains constant at \(0^\circ C\) for \(336 \, \text{min}\).
3. **From C to D**: The temperature increases linearly from \(0^\circ C\) to \(100^\circ C\) over \(420 \, \text{min}\).
4. **From D to E**: The temperature remains constant at \(100^\circ C\) for \(2268 \, \text{min}\).

### Final Graph
- The x-axis represents time (in minutes) and the y-axis represents temperature (in degrees Celsius).
- The graph will have a slope for the heating phases and flat lines for the phase change periods.