A certain amount of ice is supplied heat at a constant rate for 7 min. For the first one minute the temperature rises uniformly with time. Then it remains constant for the next 4 min and again the temperature rises at uniform rate for the last 2 min. Given `S_("ice")=0.5cal//g^(@)C,L_(f)=80cal//g`.
Final temperature at the end of 7 min is

Options

A. `10^(@)C`
B. `20^(@)C`
C. `30^(@)C`
D. `40^(@)C`

Answer

To solve the problem, we need to analyze the heat transfer process that occurs over the 7 minutes. We will break it down into three phases as described in the question.

### Step-by-Step Solution:

1. **Identify the Phases of Heating:**
- **Phase 1 (0 to 1 minute):** The temperature of ice rises uniformly.
- **Phase 2 (1 to 5 minutes):** The temperature remains constant (melting phase).
- **Phase 3 (5 to 7 minutes):** The temperature rises uniformly again.

2. **Define Variables:**
- Let \( m \) be the mass of the ice.
- Let \( Q \) be the rate of heat supplied (in calories per minute).
- Let \( \theta_i \) be the initial temperature of the ice.
- Let \( \theta_f \) be the final temperature of the water after heating.

3. **Phase 1: Heating the Ice (0 to 1 minute)**
- Heat supplied in this phase is \( Q \times 1 \).
- The temperature change is given by:
\[ Q \times 1 = m \times S_{\text{ice}} \times (\theta_f - \theta_i) \]
- This can be expressed as:
\[ Q = m \times 0.5 \times (\theta_f - \theta_i) \tag{1} \]

4. **Phase 2: Melting the Ice (1 to 5 minutes)**
- Heat supplied in this phase is \( Q \times 4 \).
- The heat required to melt the ice is:
\[ Q \times 4 = m \times L_f \]
- Substituting the latent heat:
\[ Q \times 4 = m \times 80 \tag{2} \]

5. **Phase 3: Heating the Water (5 to 7 minutes)**
- Heat supplied in this phase is \( Q \times 2 \).
- The temperature change for the water is given by:
\[ Q \times 2 = m \times S_{\text{water}} \times (\theta_f - 0) \]
- Since \( S_{\text{water}} = 1 \):
\[ Q \times 2 = m \times \theta_f \tag{3} \]

6. **Solving the Equations:**
- From Equation (1):
\[ Q = \frac{m \times 0.5 \times (\theta_f - \theta_i)}{1} \]
- From Equation (2):
\[ Q = \frac{m \times 80}{4} = 20m \]
- Setting these equal gives:
\[ 20m = 0.5m(\theta_f - \theta_i) \]
- Cancel \( m \) (assuming \( m \neq 0 \)):
\[ 20 = 0.5(\theta_f - \theta_i) \]
- Rearranging gives:
\[ \theta_f - \theta_i = 40 \tag{4} \]

7. **Substituting into Equation (3):**
- From Equation (3):
\[ Q = \frac{m \times \theta_f}{2} \]
- Setting \( Q = 20m \) gives:
\[ 20m = \frac{m \times \theta_f}{2} \]
- Cancel \( m \):
\[ 20 = \frac{\theta_f}{2} \]
- Thus:
\[ \theta_f = 40 \text{ °C} \]

### Final Answer:
The final temperature at the end of 7 minutes is **40 °C**.