Problem 67 - Master Resource Book in JEE Main Physics

Question

A room is maintained at \(20^{\circ} \mathrm{C}\) by a heater of resistance \(20 \Omega\) connected to \(200 \mathrm{~V}\) mains. The temperature is uniform throughout the room and heat is transmitted through a glass window of area \(1 \mathrm{~m}^{2}\) and thickness \(0.2 \mathrm{~cm}\). What will be the temperature outside? Given that thermal conductivity \(K\) for glass is \(0.2 \mathrm{cal} / \mathrm{m} /{ }^{\circ} \mathrm{C}\) sec and \(J=4.2 \mathrm{~J} / \mathrm{cal}\) (a) \(15.24^{\circ} \mathrm{C}\) (b) \(15.00^{\circ} \mathrm{C}\) (c) \(24.15^{\circ} \mathrm{C}\) (d) None of these

Step-by-Step Solution

Answer

The outside temperature is approximately 15.24°C (option a).

Step 1: Convert Electric Power to Heat Power

First, calculate the electric power consumed by the heater using the formula \( P = \frac{V^2}{R} \), where \( V = 200 \mathrm{~V} \) is the voltage and \( R = 20 \Omega \) is the resistance. Substitute the values:\[ P = \frac{200^2}{20} = 2000 \mathrm{~W} \]Convert watts to calories per second (since 1 W = 1 J/s and 1 cal = 4.2 J):\[ \text{Power in cal/s} = \frac{2000}{4.2} \approx 476.19 \mathrm{~cal/s} \]

Step 2: Apply Heat Conduction Formula

To maintain a constant temperature in the room, heat conducted through the glass must equal heat power from the heater. Use the heat conduction formula:\[ H = \frac{KA(T_r - T_o)}{d} \]Where:- \( H \) is the heat per second (476.19 cal/s),- \( K \) is the thermal conductivity (0.2 cal/m/°C/s),- \( A \) is the area (1 m²),- \( T_r \) is the room temperature (20°C),- \( T_o \) is the outside temperature (unknown),- \( d \) is the thickness of the glass in meters (0.2 cm or 0.002 m).

Step 3: Set Up the Heat Balance Equation

Substitute known values into the formula:\[ 476.19 = \frac{0.2 \times 1 \times (20 - T_o)}{0.002} \]

Step 4: Solve for the Outside Temperature

Rearrange and solve for \( T_o \):\[ 476.19 \times 0.002 = 0.2 \times (20 - T_o) \]\[ 0.95238 = 4 - 0.2T_o \]\[ 0.2T_o = 4 - 0.95238 \]\[ 0.2T_o = 3.04762 \]\[ T_o = \frac{3.04762}{0.2} = 15.2381 \approx 15.24^\circ \mathrm{C} \]

Step 5: Select the Correct Option

The calculated outside temperature \( 15.24^{\circ} \mathrm{C} \) matches option (a). Therefore, the correct answer is (a) \( 15.24^{\circ} \mathrm{C} \).

Key Concepts

Understanding Thermal Conductivity

Thermal conductivity is a critical concept when discussing heat transfer through materials. It measures a material's ability to conduct heat.
Every material has a unique thermal conductivity coefficient, denoted as \( K \), typically measured in \( \mathrm{cal} / \mathrm{m} /{ }^{\circ} \mathrm{C}\) per second. This coefficient outlines how much heat can pass through a unit thickness of the material when a unit temperature difference is applied.

• High thermal conductivity materials (like metals) efficiently transfer heat.
• Low thermal conductivity materials (like glass) act as insulators, reducing heat transfer.

In our exercise, the glass window has a thermal conductivity of \(0.2 \mathrm{cal} / \mathrm{m} /{ }^{\circ} \mathrm{C}\) per second. This low value suggests that the glass is not a good conductor of heat, which is important when calculating how much heat is conducted through the window with a given temperature difference.

Electric Power and Its Calculation

Electric power determines how much energy is transformed into heat by an electrical device, such as a heater. It is calculated using the formula \( P = \frac{V^2}{R} \), where \( V \) is voltage and \( R \) is resistance.
In our scenario:

• The heater operates at 200 volts.
• The resistance is 20 ohms.

By substituting these values into the formula, the power is calculated to be \( 2000 \mathrm{~W} \). This reveals how much energy per second is consumed by the heater and later converted to heat, essential for maintaining the room temperature. To relate this measure to our heat conduction context, we convert watts to calories per second since thermal conductivity is also given in calorie units. Understanding this conversion is crucial for tackling heat transfer problems effectively.

Decoding Temperature Difference and Its Role

The temperature difference is a key driver of heat transfer. When two areas are at different temperatures, heat flows naturally from the warmer region to the cooler one until equilibrium is reached.

• In our example, the room is maintained at \(20^{\circ} \mathrm{C}\).
• The outside temperature is unknown and is what we solve for.

Heat conduction through the glass window depends on this temperature difference. The heat conduction formula \( H = \frac{KA(T_r - T_o)}{d} \) uses the difference between the room temperature \( T_r \) and the outside temperature \( T_o \) to determine the rate of heat transfer through the glass.
Solving this equation allows us to find the exact temperature outside, showing how intertwined heat transfer and temperature difference are. This understanding is vital for solving practical heating and insulation problems.