250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cool from `60^@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. `1.30`
B. `0.67`
C. `0.62`
D. `0.985`

Explanation

To solve the problem, we need to calculate the specific heat of alcohol using the heat loss equations for both water and alcohol in a calorimeter. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the given data
- Mass of water (m_w) = 250 g
- Mass of alcohol (m_a) = 200 g
- Water equivalent of calorimeter (m_c) = 10 g
- Initial temperature (T_initial) = 60°C
- Final temperature (T_final) = 55°C
- Time taken for water to cool (t_w) = 130 s
- Time taken for alcohol to cool (t_a) = 67 s

### Step 2: Calculate the temperature change (ΔT)
The temperature change (ΔT) for both substances is:
\[ \Delta T = T_{initial} - T_{final} = 60°C - 55°C = 5°C \]

### Step 3: Calculate the heat loss by water and calorimeter
The heat loss (Q) can be calculated using the formula:
\[ Q = m \cdot s \cdot \Delta T \]
For water and the calorimeter:
- Total mass (m_total) = m_w + m_c = 250 g + 10 g = 260 g
- Specific heat of water (s_w) = 1 cal/g°C

Thus, the heat loss by water and calorimeter in 130 seconds is:
\[ Q_w = (m_w + m_c) \cdot s_w \cdot \Delta T = 260 \cdot 1 \cdot 5 = 1300 \text{ cal} \]
Since this occurs over 130 seconds, the rate of heat loss is:
\[ \text{Rate of heat loss by water} = \frac{1300 \text{ cal}}{130 \text{ s}} = 10 \text{ cal/s} \]

### Step 4: Calculate the heat loss by alcohol and calorimeter
For alcohol:
- Total mass = m_a + m_c = 200 g + 10 g = 210 g
- Let the specific heat of alcohol be \( s_a \).

The heat loss by alcohol and calorimeter in 67 seconds is:
\[ Q_a = (m_a + m_c) \cdot s_a \cdot \Delta T = 210 \cdot s_a \cdot 5 \]
Since this occurs over 67 seconds, the rate of heat loss is:
\[ \text{Rate of heat loss by alcohol} = \frac{210 \cdot s_a \cdot 5}{67} \]

### Step 5: Set the heat loss equations equal
Since the heat loss rates are equal:
\[ 10 = \frac{210 \cdot s_a \cdot 5}{67} \]

### Step 6: Solve for the specific heat of alcohol (s_a)
Rearranging the equation gives:
\[ 10 \cdot 67 = 210 \cdot s_a \cdot 5 \]
\[ 670 = 1050 \cdot s_a \]
\[ s_a = \frac{670}{1050} = \frac{67}{105} \approx 0.6381 \text{ cal/g°C} \]

### Step 7: Final answer
The specific heat of alcohol is approximately:
\[ s_a \approx 0.638 \text{ cal/g°C} \quad (\text{rounded to } 0.62 \text{ cal/g°C}) \]

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

---