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 step-by-step, we will follow these calculations:

### Step 1: Calculate the power output of the heater
The power \( P \) produced 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 using the mechanical equivalent of heat \( 1 \text{ cal} = 4.2 \text{ J} \), we convert the power to calories per second:
\[ P = \frac{2000 \text{ J/s}}{4.2 \text{ J/cal}} \approx 476.2 \text{ cal/s} \]

### Step 3: Set up the heat transfer equation through the glass window
The heat transfer \( \frac{dQ}{dt} \) through the glass window can be calculated using Fourier's law of heat conduction:
\[ \frac{dQ}{dt} = \frac{KA(T_1 - T_2)}{L} \]
where:
- \( K = 0.2 \text{ cal/m°C} \) (thermal conductivity of glass)
- \( A = 1 \text{ m}^2 \) (area of the window)
- \( T_1 = 20 \text{ °C} \) (inside temperature)
- \( T_2 = \theta \) (outside temperature)
- \( L = 0.2 \text{ cm} = 0.002 \text{ m} \) (thickness of the glass)

Substituting the values:
\[ \frac{dQ}{dt} = \frac{0.2 \times 1 \times (20 - \theta)}{0.002} = 100(20 - \theta) \text{ cal/s} \]

### Step 4: Equate the heat produced by the heater to the heat lost through the window
Since the temperature in the room remains constant, the heat produced by the heater equals the heat lost through the window:
\[ 476.2 = 100(20 - \theta) \]

### Step 5: Solve for the outside temperature \( \theta \)
Expanding the equation:
\[ 476.2 = 2000 - 100\theta \]
Rearranging gives:
\[ 100\theta = 2000 - 476.2 \]
\[ 100\theta = 1523.8 \]
\[ \theta = \frac{1523.8}{100} = 15.238 \text{ °C} \]

### Final Answer
The outside temperature is approximately:
\[ \theta \approx 15.24 \text{ °C} \]

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