60 g of ice at `0^(@)C` is added to 200 g of water initially at `70^(@)C` in a calorimeter of unknown water equivalent W. If the final temperature of the mixture is `40^(@)C`, then the value of W is [Take latent heat of fusion of ice `L_(f) = 80 cal g^(-1)` and specific heat capacity of water `s = 1 cal g^(-1).^(@)C^(-1)`]

Options

A. 70 g
B. 80 g
C. 40 g
D. 20 g

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat required to melt the ice and raise its temperature
1. **Calculate the heat required to melt the ice (Q1)**:
- Mass of ice (m_ice) = 60 g
- Latent heat of fusion (L_f) = 80 cal/g
- Heat required to melt the ice:
\[ Q_1 = m_{\text{ice}} \times L_f = 60 \, \text{g} \times 80 \, \text{cal/g} = 4800 \, \text{cal} \]

2. **Calculate the heat required to raise the temperature of the melted ice from 0°C to 40°C (Q2)**:
- Mass of water formed from melted ice = 60 g
- Specific heat capacity of water (s) = 1 cal/g°C
- Temperature change (ΔT) = 40°C - 0°C = 40°C
- Heat required to raise the temperature:
\[ Q_2 = m_{\text{water}} \times s \times \Delta T = 60 \, \text{g} \times 1 \, \text{cal/g°C} \times 40 \, \text{°C} = 2400 \, \text{cal} \]

3. **Total heat required (Q_total)**:
\[ Q_{\text{total}} = Q_1 + Q_2 = 4800 \, \text{cal} + 2400 \, \text{cal} = 7200 \, \text{cal} \]

### Step 2: Calculate the heat released by the warmer water and the calorimeter
1. **Calculate the heat released by the water and the calorimeter**:
- Mass of water (m_water) = 200 g
- Water equivalent of the calorimeter (W) = unknown
- Initial temperature of water = 70°C
- Final temperature of mixture = 40°C
- Temperature change (ΔT) = 70°C - 40°C = 30°C
- Heat released:
\[ Q_{\text{released}} = (m_{\text{water}} + W) \times s \times \Delta T = (200 \, \text{g} + W) \times 1 \, \text{cal/g°C} \times 30 \, \text{°C} \]
\[ Q_{\text{released}} = (200 + W) \times 30 \, \text{cal} \]

### Step 3: Set the heat required equal to the heat released
1. **Equate the heat required to the heat released**:
\[ 7200 \, \text{cal} = (200 + W) \times 30 \, \text{cal} \]

2. **Solve for W**:
\[ 7200 = 6000 + 30W \]
\[ 7200 - 6000 = 30W \]
\[ 1200 = 30W \]
\[ W = \frac{1200}{30} = 40 \, \text{g} \]

### Final Answer:
The water equivalent of the calorimeter (W) is **40 g**.

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