250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same colorimeter and cool from `606^@C` to `55^@C` in 130 s and 67 s, respectively. If the water equivalent of the calorimeter is 10 g, then the specific heat of alcohol in `cal//g^@C` is

Options

A. 1.3
B. 0.67
C. 0.62
D. 0.985

Explanation

To find the specific heat of alcohol, we can use the principle of calorimetry, which states that the heat lost by the warmer substance (water) is equal to the heat gained by the cooler substance (alcohol and calorimeter).

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of water, \( m_w = 250 \, \text{g} \)
- Mass of alcohol, \( m_a = 200 \, \text{g} \)
- Water equivalent of the calorimeter, \( W = 10 \, \text{g} \)
- Initial temperature, \( T_i = 60 \, \text{°C} \)
- Final temperature, \( T_f = 55 \, \text{°C} \)
- Time for water to cool, \( t_w = 130 \, \text{s} \)
- Time for alcohol to cool, \( t_a = 67 \, \text{s} \)

2. **Calculate the Temperature Change (\( \Delta T \)):**
\[ \Delta T = T_i - T_f = 60 \, \text{°C} - 55 \, \text{°C} = 5 \, \text{°C} \]

3. **Calculate Heat Lost by Water:**
The heat lost by water can be calculated using the formula:
\[ Q_w = m_w \cdot s_w \cdot \Delta T + W \cdot \Delta T \]
where \( s_w = 1 \, \text{cal/g°C} \) (specific heat of water).
\[ Q_w = 250 \cdot 1 \cdot 5 + 10 \cdot 5 = 1250 + 50 = 1300 \, \text{cal} \]

4. **Calculate Rate of Heat Loss for Water:**
\[ \text{Rate of heat loss for water} = \frac{Q_w}{t_w} = \frac{1300}{130} = 10 \, \text{cal/s} \]

5. **Calculate Heat Lost by Alcohol:**
The heat lost by alcohol can be expressed as:
\[ Q_a = m_a \cdot s_a \cdot \Delta T + W \cdot \Delta T \]
where \( s_a \) is the specific heat of alcohol.
\[ Q_a = 200 \cdot s_a \cdot 5 + 10 \cdot 5 = 1000s_a + 50 \]

6. **Calculate Rate of Heat Loss for Alcohol:**
\[ \text{Rate of heat loss for alcohol} = \frac{Q_a}{t_a} = \frac{1000s_a + 50}{67} \]

7. **Set the Rates Equal:**
Since the rate of heat loss is the same for both substances:
\[ 10 = \frac{1000s_a + 50}{67} \]

8. **Solve for \( s_a \):**
Multiply both sides by 67:
\[ 670 = 1000s_a + 50 \]
Rearranging gives:
\[ 1000s_a = 670 - 50 = 620 \]
\[ s_a = \frac{620}{1000} = 0.62 \, \text{cal/g°C} \]

### Final Answer:
The specific heat of alcohol is \( 0.62 \, \text{cal/g°C} \).