Question

The inner and outer surfaces of a \(0.5-\mathrm{cm}\) thick $2-\mathrm{m} \times 2-\mathrm{m}\( window glass in winter are \)10^{\circ} \mathrm{C}$ and \(3^{\circ} \mathrm{C}\), respectively. If the thermal conductivity of the glass is \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the amount of heat loss through the glass over a period of \(5 \mathrm{~h}\). What would your answer be if the glass were 1 cm thick?

Short answer

Answer: The heat loss through the 0.5 cm thick glass over 5 hours is 78,585,600 J, and the heat loss through the 1 cm thick glass over the same time period is 39,292,800 J.

Step-by-step solution

Write down the given information

The dimensions of the window glass are 2 m × 2 m and the thickness is 0.5 cm. The inner surface temperature is \(T_1 = 10^{\circ} \mathrm{C}\), and the outer surface temperature is \(T_2 = 3^{\circ} \mathrm{C}\). The thermal conductivity of the glass is \(k = 0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and we need to calculate the heat loss for a period of 5 h.

Calculate the temperature difference

The temperature difference between the inner and outer surfaces of the glass is: \(\Delta T = T_1 - T_2 = 10^{\circ} \mathrm{C} - 3^{\circ} \mathrm{C} = 7 \mathrm{K}\).

Convert the thickness to meters

The thickness of the glass is given in centimeters, so we need to convert it to meters: \(d = 0.5 \mathrm{cm} \times \dfrac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.005 \mathrm{m}\).

Calculate the rate of heat transfer

Using the formula for heat transfer through conduction, we find the rate of heat transfer, \(q\): \(q = \dfrac{k \cdot A \cdot \Delta T}{d}\), where \(A\) is the area of the window glass. We have \(A = 2 \mathrm{m} \times 2 \mathrm{m} = 4 \mathrm{m^2}\), so \(q = \dfrac{0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 4 \mathrm{m^2} \cdot 7 \mathrm{K}}{0.005 \mathrm{m}} = 4368 \mathrm{W}\).

Determine the heat loss for 5 hours

To find the total heat loss for a period of 5 hours, we multiply the rate of heat transfer by the time: \(Q = q \cdot t = 4368 \mathrm{W} \cdot 5 \mathrm{h} \cdot \dfrac{3600 \mathrm{s}}{1 \mathrm{h}} = 78585600 \mathrm{J}\).

Repeat the calculation for 1 cm thickness

Now, we have to repeat the calculation with a thickness of 1 cm, which is 0.01 m: \(d' = 1 \mathrm{cm} \times \dfrac{1 \mathrm{m}}{100 \mathrm{cm}} = 0.01 \mathrm{m}\). \(q' = \dfrac{k \cdot A \cdot \Delta T}{d'} = \dfrac{0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 4 \mathrm{m^2} \cdot 7 \mathrm{K}}{0.01 \mathrm{m}} = 2184 \mathrm{W}\). \(Q' = q' \cdot t = 2184 \mathrm{W} \cdot 5 \mathrm{h} \cdot \dfrac{3600 \mathrm{s}}{1 \mathrm{h}} = 39292800 \mathrm{J}\).

Write the final answers

The heat loss through the 0.5 cm thick glass over 5 hours is \(Q = 78585600 \mathrm{J}\), and the heat loss through the 1 cm thick glass over the same time period is \(Q' = 39292800 \mathrm{J}\).