A room is maintained at 20°C by a heater of resistance 20 ohm connected to 200 V mains. The temperat

Q: Why do we equate electrical power to the rate of heat conduction?

Because the room is maintained at a constant temperature. This means it's in a steady state where energy in equals energy out.

Q: Why did we multiply $K$ by 4.2?

The thermal conductivity $K$ was given in calories per meter per second per degree Celsius. Since our electrical power is in Watts (Joules per second), we must convert calories to Joules using the mechanical equivalent of heat ($1\ \text{cal} = 4.2\ \text{J}$).

Q: Does the area of the walls matter?

The problem states that heat is transmitted *through a glass window*. We assume the rest of the room is perfectly insulated, so heat only escapes through the window.

Q: What if the outside temperature was higher than the inside?

Then heat would flow *into* the room, and the heater would not be needed to maintain the temperature. In fact, an air conditioner would be required!

Q: Why did we convert the thickness to meters?

To maintain consistency in SI units. Since $K$ is given per meter and area is in square meters, the thickness must also be in meters to cancel the units correctly.

Question

A room is maintained at $20^{\circ}	ext{C}$ by a heater of resistance $20\ \Omega$ connected to $200\ 	ext{V}$ mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area $1\ 	ext{m}^2$ and thickness $0.2\ 	ext{cm}$. Calculate the temperature outside. Thermal conductivity of glass is $0.2\ 	ext{cal m}^{-1}	ext{s}^{-1} (^{\circ}	ext{C})^{-1}$ and mechanical equivalent of heat is $4.2\ 	ext{Jcal}^{-1}$.

Sigma Percentile · Accepted Answer

### The Concept of Steady State

Imagine a cozy room on a freezing winter night. The heater is working hard to keep the room at a comfortable $20^\circ	ext{C}$. But the heat doesn't just stay in the room; it constantly escapes through the glass window.

For the room's temperature to remain perfectly constant, a delicate balance must be struck. The rate at which the heater produces thermal energy must exactly equal the rate at which heat is conducted away through the window. This beautiful equilibrium is known as the **steady state**.

### Calculating the Heater's Power

First, let's determine how much heat our electrical heater is pumping into the room every second. We are given the voltage $V = 200\ 	ext{V}$ and the resistance $R = 20\ \Omega$.

Using the electrical power formula, we can find the rate of heat generation:
$$P = \frac{V^2}{R}$$

Substituting the given values:
$$P = \frac{(200)^2}{20} = \frac{40000}{20} = 2000\ 	ext{W}$$

So, the heater is generating $2000\ 	ext{Joules}$ of heat every single second.

### The Law of Heat Conduction

Now, let's look at the heat escaping through the window. According to **Fourier's Law of Heat Conduction**, the rate of heat flow $\frac{dQ}{dt}$ through a material is given by:
$$\frac{dQ}{dt} = \frac{KA(	heta_1 - 	heta_2)}{l}$$

Here, $K$ is the thermal conductivity, $A$ is the area, $	heta_1 - 	heta_2$ is the temperature difference, and $l$ is the thickness of the material.

### The Unit Conversion Trap

Before we plug in the numbers, we must be extremely careful with our units. This is a classic trap! The thermal conductivity $K$ is given as $0.2\ 	ext{cal m}^{-1}	ext{s}^{-1} (^\circ	ext{C})^{-1}$.

Since our heater's power is in Watts (Joules per second), we must convert calories to Joules. We use the mechanical equivalent of heat, where $1\ 	ext{cal} = 4.2\ 	ext{J}$.
$$K = 0.2 	imes 4.2 = 0.84\ 	ext{W m}^{-1}(^\circ	ext{C})^{-1}$$

Also, the thickness of the glass is given in centimeters. We must convert it to meters to match the other SI units:
$$l = 0.2\ 	ext{cm} = 0.002\ 	ext{m}$$

### The Final Calculation

Now, we equate the rate of heat generation to the rate of heat loss:
$$\frac{V^2}{R} = \frac{KA(20 - 	heta)}{l}$$

Substituting all our carefully converted values:
$$2000 = \frac{0.84 	imes 1 	imes (20 - 	heta)}{0.002}$$

Let's simplify the right side of the equation. Dividing $0.84$ by $0.002$ gives $420$:
$$2000 = 420(20 - 	heta)$$

Now, we isolate the temperature difference:
$$20 - 	heta = \frac{2000}{420} \approx 4.76^\circ	ext{C}$$

Finally, solving for the outside temperature $	heta$:
$$	heta = 20 - 4.76 = 15.24^\circ	ext{C}$$

The outside temperature is $15.24^\circ	ext{C}$. The physics of steady state perfectly balances the electrical power and thermal conduction!

Enter Numerical Value:

Visualized Solution

$Steady State Heat Transfer$

$P = \frac{V ^{2}}{R}$

$\frac{d Q}{d t} = \frac{K A ( θ _{1} - θ _{2} )}{l}$

$K = 0.2 \times 4.2 J m^{- 1} s^{- 1} (^{\circ} C)^{- 1}$

$\frac{V ^{2}}{R} = \frac{K A ( 20 - θ )}{l}$

$2000 = \frac{0.84 ( 20 - θ )}{0.002}$

$θ = 15.2 4^{\circ} C$

$What if the window was double-paned?$

The Sigma Insight: Heat Transfer

The Concept of Steady State

Calculating the Heater's Power

The Law of Heat Conduction

The Unit Conversion Trap

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Steady State Heat Transfer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=RV2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdQ​=lKA(θ1​−θ2​)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K=0.2×4.2 J m−1s−1(∘C)−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RV2​=lKA(20−θ)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2000=0.0020.84(20−θ)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=15.24∘C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the window was double-paned?