An immersion heater, in an insulated vessel of negligible heat capacity brings 100 g of water to the boiling point from `16^@C` in 7 min. Then
Q. The water is replaced by 200 g of alcohol, which is heated from `16^@C` to the boiling point of `78^@C` in 6 min 12 s whereas 30 g are vapourized in 5 min 6 s. The specific heat of alcohol is

Options

A. `0.6 J//kg^@C`
B. `0.6 cal//g^@C`
C. `0.6 cal//kg^@C`
D. `6 J//kg^@C`

Explanation

To solve the problem step by step, we will calculate the specific heat of alcohol using the information given in the question.

### Step 1: Calculate the heat required to bring water to boiling point

The heat \( Q_W \) required to bring 100 g of water from \( 16^\circ C \) to \( 100^\circ C \) can be calculated using the formula:

\[ Q_W = m \cdot c \cdot \Delta T \]

Where:
- \( m = 100 \, \text{g} \) (mass of water)
- \( c = 1 \, \text{cal/g}^\circ C \) (specific heat of water)
- \( \Delta T = 100^\circ C - 16^\circ C = 84^\circ C \)

Substituting the values:

\[ Q_W = 100 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot 84^\circ C = 8400 \, \text{cal} \]

### Step 2: Calculate the power of the immersion heater

The power \( P \) of the immersion heater can be calculated using the formula:

\[ P = \frac{Q_W}{t} \]

Where \( t \) is the time in seconds. Given that the time is 7 minutes:

\[ t = 7 \, \text{min} \times 60 \, \text{s/min} = 420 \, \text{s} \]

Now substituting the values:

\[ P = \frac{8400 \, \text{cal}}{420 \, \text{s}} = 20 \, \text{cal/s} \]

To convert this to watts (1 cal/s = 4.184 W):

\[ P = 20 \, \text{cal/s} \times 4.184 \, \text{W/cal} = 83.68 \, \text{W} \approx 84 \, \text{W} \]

### Step 3: Calculate the heat required to heat alcohol

Now, we replace the water with 200 g of alcohol and heat it from \( 16^\circ C \) to \( 78^\circ C \). The temperature change \( \Delta T \) for alcohol is:

\[ \Delta T = 78^\circ C - 16^\circ C = 62^\circ C \]

The heat \( Q_A \) required for the alcohol can be expressed as:

\[ Q_A = m \cdot C_A \cdot \Delta T \]

Where:
- \( m = 200 \, \text{g} \) (mass of alcohol)
- \( C_A \) is the specific heat of alcohol (unknown)
- \( \Delta T = 62^\circ C \)

### Step 4: Calculate the time for heating alcohol

The time taken to heat the alcohol is given as 6 minutes 12 seconds:

\[ t = 6 \, \text{min} \times 60 \, \text{s/min} + 12 \, \text{s} = 372 \, \text{s} \]

The heat \( Q_A \) can also be expressed in terms of power:

\[ Q_A = P \cdot t = 84 \, \text{W} \cdot 372 \, \text{s} = 31248 \, \text{J} \]

Converting this to calories (1 cal = 4.184 J):

\[ Q_A = \frac{31248 \, \text{J}}{4.184 \, \text{J/cal}} \approx 7460 \, \text{cal} \]

### Step 5: Set up the equation for specific heat of alcohol

Now we can set up the equation:

\[ 200 \, \text{g} \cdot C_A \cdot 62^\circ C = 7460 \, \text{cal} \]

Solving for \( C_A \):

\[ C_A = \frac{7460 \, \text{cal}}{200 \, \text{g} \cdot 62^\circ C} = \frac{7460}{12400} \approx 0.6 \, \text{cal/g}^\circ C \]

### Final Answer

The specific heat of alcohol is approximately \( 0.6 \, \text{cal/g}^\circ C \).

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