A room at `20^@C` is heated by a heater of resistence 20 ohm connected to 200 VV mains. The temperature is uniform throughout the room and the heati s transmitted through a glass window of area `1m^2` and thickness 0.2 cm. Calculate the temperature outside. Thermal conductivity of glass is `0.2 cal//mC^@` s and mechanical equivalent of heat is `4.2 J//cal`.

Options

A. `13.69^@C`
B. `15.24^@C`
C. `17.85^@C`
D. `19.96^@C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the power generated by the heater.
The power (P) generated by the heater can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage (200 V) and \( R \) is the resistance (20 ohms).

Substituting the values:
\[ P = \frac{200^2}{20} = \frac{40000}{20} = 2000 \text{ W} \]

### Step 2: Convert power from watts to calories per second.
Since \( 1 \text{ W} = \frac{1 \text{ J}}{1 \text{s}} \) and \( 1 \text{ cal} = 4.2 \text{ J} \), we can convert the power:
\[ P = \frac{2000 \text{ J/s}}{4.2 \text{ J/cal}} \approx 476.19 \text{ cal/s} \]

### Step 3: Write the heat transfer equation through the glass.
The rate of heat transfer (Q) through the glass can be expressed using Fourier's law:
\[ Q = \frac{k \cdot A \cdot \Delta T}{d} \]
where:
- \( k = 0.2 \text{ cal/m°C·s} \) (thermal conductivity of glass)
- \( A = 1 \text{ m}^2 \) (area of the window)
- \( \Delta T = 20 - T_0 \) (temperature difference)
- \( d = 0.2 \text{ cm} = 0.002 \text{ m} \) (thickness of the glass)

### Step 4: Substitute the known values into the heat transfer equation.
Substituting the values into the equation:
\[ 476.19 = \frac{0.2 \cdot 1 \cdot (20 - T_0)}{0.002} \]

### Step 5: Simplify the equation.
Multiplying both sides by \( 0.002 \):
\[ 476.19 \cdot 0.002 = 0.2 \cdot (20 - T_0) \]
\[ 0.95238 = 0.2 \cdot (20 - T_0) \]

### Step 6: Solve for \( T_0 \).
Dividing both sides by 0.2:
\[ \frac{0.95238}{0.2} = 20 - T_0 \]
\[ 4.7619 = 20 - T_0 \]
Rearranging gives:
\[ T_0 = 20 - 4.7619 \approx 15.2381 \text{ °C} \]

### Final Answer:
The temperature outside is approximately \( 15.24 \text{ °C} \).

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