A mass of 50g of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from `30^(@)C` to `25^(@)C`. A mass of 100g of another liquid in an identical vessel with identical surroundings takes the same time to cool from `30^(@)`C to `25^(@)C`. The specific heat of the liquid is: (The water equivalent of the vessel is 30g.)

Options

A. 2.0 kcal/kg
B. 7 kcal/kg
C. 3kcal/kg
D. 0.5 kcal/kg

Answer

To solve the problem, we will use the principle of heat transfer and Newton's law of cooling. We will analyze the cooling of both the water and the other liquid in terms of their heat loss.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Mass of water (M1) = 50 g
- Initial temperature of water (T_initial) = 30°C
- Final temperature of water (T_final) = 25°C
- Mass of the other liquid (M2) = 100 g
- Water equivalent of the vessel (W) = 30 g
- Time taken for cooling (t) = 2 minutes (but will not affect the final calculation as it cancels out)

2. **Calculate the Temperature Change (ΔT):**
- For both substances, the temperature change (ΔT) is the same:
\[ \Delta T = T_{initial} - T_{final} = 30°C - 25°C = 5°C \]

3. **Apply the Heat Loss Formula:**
- According to Newton's law of cooling, the heat lost by the water can be expressed as:
\[ Q_{water} = M_1 \cdot C_{water} \cdot \Delta T + W \cdot \Delta T \]
- Where \(C_{water} = 1 \, \text{cal/g°C}\) (specific heat of water).
- Substitute the values:
\[ Q_{water} = 50 \cdot 1 \cdot 5 + 30 \cdot 5 = 250 + 150 = 400 \, \text{cal} \]

4. **Set Up the Equation for the Other Liquid:**
- For the other liquid, the heat lost can be expressed as:
\[ Q_{liquid} = M_2 \cdot C_{liquid} \cdot \Delta T + W \cdot \Delta T \]
- Substitute the values:
\[ Q_{liquid} = 100 \cdot C_{liquid} \cdot 5 + 30 \cdot 5 \]
- This simplifies to:
\[ Q_{liquid} = 500 \cdot C_{liquid} + 150 \]

5. **Equate the Heat Losses:**
- Since both processes take the same time and occur under identical conditions, we can equate the heat lost:
\[ Q_{water} = Q_{liquid} \]
\[ 400 = 500 \cdot C_{liquid} + 150 \]

6. **Solve for the Specific Heat of the Liquid (C_liquid):**
- Rearranging the equation gives:
\[ 500 \cdot C_{liquid} = 400 - 150 \]
\[ 500 \cdot C_{liquid} = 250 \]
\[ C_{liquid} = \frac{250}{500} = 0.5 \, \text{cal/g°C} \]

### Final Answer:
The specific heat of the liquid is \(0.5 \, \text{cal/g°C}\).