A glass window pane of thickness 4.0\text{ mm} and surface area 1.5\text{ m}^2 maintains a… — JAMB UTME

A glass window pane of thickness $4.0\text{ mm}$ and surface area $1.5\text{ m}^2$ maintains an inner surface temperature of $20^\circ\text{C}$ and an outer surface temperature of $5^\circ\text{C}$ . If the thermal conductivity of glass is $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , what is the rate of heat transfer by conduction through the window?

$4500\text{ W}$ Answer
B
$4.5\text{ W}$
C
$86,400\text{ W}$
D
$0.072\text{ W}$

Answer

4500\text{ W}

(or

4.5\text{ kW}

)

The rate of conductive heat transfer is governed by Fourier's law of thermal conduction:

\frac{Q}{t} = \frac{k A (T_1 - T_2)}{d}

. Substituting

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

A = 1.5\text{ m}^2

, temperature difference

\Delta T = 15\text{ K}

, and thickness

d = 0.004\text{ m}

gives

\frac{Q}{t} = \frac{0.80 \times 1.5 \times 15}{0.004} = 4500\text{ W}

Step-by-Step Solution

Identify the given parameters and convert units to standard SI units

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

A = 1.5\text{ m}^2

\Delta T = 20^\circ\text{C} - 5^\circ\text{C} = 15\text{ K}

, and

d = 4.0\text{ mm} = 4.0 \times 10^{-3}\text{ m}

Thermal conductivity formulas require distance/thickness in meters.

Apply the law of thermal conduction formula

\frac{Q}{t} = \frac{k A \Delta T}{d}

The rate of heat transfer by conduction is directly proportional to thermal conductivity, surface area, and temperature difference, and inversely proportional to thickness.

Substitute the values and calculate the heat transfer rate

\frac{Q}{t} = \frac{0.80 \times 1.5 \times 15}{4.0 \times 10^{-3}} = \frac{18}{0.004} = 4500\text{ W}

Performing accurate arithmetic yields

4500\text{ Joules per second}

(Watts).

Key Concept

Rate of thermal conduction through a uniform slab

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