The temperature of a body rises by `44^(@)C` when a certain amount of heat is given to it . The same heat when supplied to `22` g of ice at `-8^(@)C` , raises its temperature by `16^(@)C`. The water eqivalent of the body is `[Given : s_(water) = 1"cal"//g^(@)C` & `L_(t) = 80 "cal" //g, s_("ice") = 0.5"cal"//g^(@)C`]

Options

A. `25`g
B. `50` g
C. `80` g
D. `100` g

Explanation

To solve the problem, we need to calculate the water equivalent of the body based on the heat transfer involved in raising the temperature of ice and the body itself.

### Step-by-Step Solution:

1. **Calculate the heat required to raise the temperature of ice:**
- The heat required to raise the temperature of ice from -8°C to 0°C:
\[ Q_1 = m \cdot s_{\text{ice}} \cdot \Delta T_1 \]
where:
- \( m = 22 \, \text{g} \) (mass of ice)
- \( s_{\text{ice}} = 0.5 \, \text{cal/g°C} \) (specific heat of ice)
- \( \Delta T_1 = 8 \, \text{°C} \) (temperature change from -8°C to 0°C)

\[ Q_1 = 22 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 8 \, \text{°C} = 88 \, \text{cal} \]

2. **Calculate the heat required to melt the ice:**
- The heat required to melt the ice at 0°C:
\[ Q_2 = m \cdot L_t \]
where:
- \( L_t = 80 \, \text{cal/g} \) (latent heat of fusion)

\[ Q_2 = 22 \, \text{g} \cdot 80 \, \text{cal/g} = 1760 \, \text{cal} \]

3. **Calculate the heat required to raise the temperature of water:**
- The heat required to raise the temperature of the resulting water from 0°C to 16°C:
\[ Q_3 = m \cdot s_{\text{water}} \cdot \Delta T_2 \]
where:
- \( s_{\text{water}} = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T_2 = 16 \, \text{°C} \) (temperature change from 0°C to 16°C)

\[ Q_3 = 22 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 16 \, \text{°C} = 352 \, \text{cal} \]

4. **Calculate the total heat absorbed by the ice:**
- The total heat \( Q \) absorbed by the ice is:
\[ Q = Q_1 + Q_2 + Q_3 = 88 \, \text{cal} + 1760 \, \text{cal} + 352 \, \text{cal} = 2200 \, \text{cal} \]

5. **Relate the heat given to the body to its temperature change:**
- The heat given to the body, which causes a temperature rise of 44°C, can be expressed as:
\[ Q = m' \cdot s_{\text{body}} \cdot \Delta T \]
where:
- \( m' \) is the water equivalent of the body,
- \( s_{\text{body}} \) is the specific heat of the body (not given, but we can use the water equivalent concept).

6. **Set the equations equal:**
- Since the same amount of heat is given to the body:
\[ 2200 \, \text{cal} = m' \cdot 1 \, \text{cal/g°C} \cdot 44 \, \text{°C} \]

7. **Solve for the water equivalent \( m' \):**
\[ m' = \frac{2200 \, \text{cal}}{44 \, \text{°C}} = 50 \, \text{g} \]

### Final Answer:
The water equivalent of the body is **50 g**.