An electric heater is used in a room of total wall area 137 m^2 to maintain a temperature of +20°C i

Q: Why are the thermal resistances of the layers added in series?

Heat flows through the wall from the inside to the outside, passing through each layer sequentially. Just like electrical current passing through resistors one after another, the thermal resistances add up.

Q: Why do we need to convert the thickness from cm to m?

The thermal conductivity $K$ is given in SI units ($\text{W/m/}^\circ\text{C}$) and the area is in $\text{m}^2$. To maintain dimensional consistency and get the resistance in standard units, the thickness must be in meters.

Q: How do we calculate the temperature difference $\Delta T$?

The temperature difference is the higher temperature minus the lower temperature. Here, it's the inside temperature ($20^\circ\text{C}$) minus the outside temperature ($-10^\circ\text{C}$), which gives $20 - (-10) = 30^\circ\text{C}$.

Q: Why is the heater power exactly equal to the rate of heat loss?

To maintain a constant temperature inside the room (steady state), the net heat flow must be zero. This means the heat added by the heater must perfectly balance the heat lost through the walls.

Q: The book's answer is 9091 W, but the exact calculation gives 9000 W. Why the difference?

The book prematurely approximated the total thermal resistance $1/300 \approx 0.00333...$ as $0.33 \times 10^{-2}$. Dividing by this rounded number gives $30 / 0.0033 \approx 9091 \text{ W}$. However, using the exact fraction $1/300$ yields exactly $9000 \text{ W}$.

Question

An electric heater is used in a room of total wall area $137 	ext{ m}^2$ to maintain a temperature of $+20^\circ	ext{C}$ inside it, when the outside temperature is $-10^\circ	ext{C}$. The walls have three different layers. The innermost layer is of wood of thickness $2.5 	ext{ cm}$, the middle layer is of cement of thickness $1.0 	ext{ cm}$ and the outermost layer is of brick of thickness $25.0 	ext{ cm}$. Find the power of the electric heater. Assume that there is no heat loss through the floor and the ceiling. The thermal conductivities of wood, cement and brick are $0.125$, $1.5$ and $1.0 	ext{ W/m/}^\circ	ext{C}$ respectively.

Sigma Percentile · Accepted Answer

### The Winter Chill and the Cozy Room

Imagine a room in the dead of winter. Inside, it is a cozy $20^\circ	ext{C}$, but outside, it is a freezing $-10^\circ	ext{C}$. The walls of this room are not just made of one material; they are constructed from three distinct layers: wood on the inside, cement in the middle, and brick on the outside.

We need to find the power of the electric heater required to maintain this temperature difference. To do this, we must understand that the heater must supply heat at the exact same rate at which it is being lost through the walls.

### The Electrical Analogy: Thermal Resistance

This is a classic case of steady-state heat conduction. The three layers of the wall act exactly like three electrical resistors connected in series. Heat flows through them sequentially, from the warm inside to the cold outside.

The formula for thermal resistance is $R = \frac{l}{KA}$, where $l$ is the thickness of the layer, $K$ is its thermal conductivity, and $A$ is the cross-sectional area. Because the layers are in series, the total equivalent thermal resistance is simply the sum of the individual resistances:

$$R_{	ext{eq}} = R_1 + R_2 + R_3$$

### Calculating the Resistance of Each Layer

Let's calculate the thermal resistance of each layer. We must be careful to convert the thickness from centimeters to meters to maintain SI units.

For the wood layer, the thickness is $2.5 	ext{ cm}$ ($2.5 	imes 10^{-2} 	ext{ m}$), the thermal conductivity is $0.125 	ext{ W/m/}^\circ	ext{C}$, and the area is $137 	ext{ m}^2$:

$$R_1 = \frac{2.5 	imes 10^{-2}}{0.125 	imes 137}$$

Similarly, for the cement and brick layers:

$$R_2 = \frac{1.0 	imes 10^{-2}}{1.5 	imes 137}$$

$$R_3 = \frac{25.0 	imes 10^{-2}}{1.0 	imes 137}$$

### The Magic of Exact Math

Now, let's add these up. If we factor out the area ($137 	ext{ m}^2$), the terms inside the bracket simplify beautifully:

$$R_{	ext{eq}} = \frac{1}{137} \left( \frac{0.025}{0.125} + \frac{0.01}{1.5} + \frac{0.25}{1.0} ight)$$

$$R_{	ext{eq}} = \frac{1}{137} \left( \frac{1}{5} + \frac{1}{150} + \frac{1}{4} ight)$$

Finding a common denominator of $150$, the numerators add up to exactly $68.5$. And beautifully, $68.5$ is exactly half of $137$!

$$R_{	ext{eq}} = \frac{1}{137} \left( \frac{30 + 1 + 37.5}{150} ight) = \frac{1}{137} \left( \frac{68.5}{150} ight) = \frac{1}{300} ^\circ	ext{C/W}$$

### The Final Power Requirement

With this exact thermal resistance, the rate of heat transfer is simply the temperature difference divided by the total resistance. The temperature difference is $20 - (-10) = 30^\circ	ext{C}$.

$$\frac{dQ}{dt} = \frac{\Delta T}{R_{	ext{eq}}} = \frac{30}{1/300} = 9000 	ext{ W}$$

**Note:** If you prematurely approximate the resistance to $0.00333$, you would get $9091 	ext{ W}$, which is what some textbooks show. But the exact, elegant answer is **$9000 	ext{ W}$**!

To keep the room warm, the electric heater must supply power at this exact same rate. Therefore, the required power of the heater is $9000 	ext{ W}$.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$The Master Equation$

$Thermal Resistance Formula$

$Resistance of Wood Layer$

$Resistance of Cement and Brick$

$Total Equivalent Resistance$

$Rate of Heat Transfer$

$Final Power Requirement$

$The Way Forward$

The Sigma Insight: Heat Transfer

The Winter Chill and the Cozy Room

The Electrical Analogy

Calculating the Resistance of Each Layer

The Magic of Exact Math

The Final Power Requirement

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Master Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermal Resistance Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance of Wood Layer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance of Cement and Brick

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Equivalent Resistance

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Power Requirement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward