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Chapter 1: Problem 54

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 \(1 \mathrm{~cm}\) thick?

Short Answer

Expert verified
Question: Calculate the heat loss through a 2m x 2m glass window for each thickness of 0.5 cm and 1 cm over a period of 5 hours if the temperature difference between the inner and outer surfaces is 7°C and the thermal conductivity of the glass is 0.78 W/m·K. Answer: The heat loss through the 0.5 cm thick glass window is 34128 Ws and through the 1 cm thick glass window is 17064 Ws over a period of 5 hours.

Step by step solution

01

Identify the known variables

The thickness of the glass (L) is 0.5 cm which needs to be converted to meters, the surface area of the window is A = 2m * 2m, and the temperature difference between the inner and outer surfaces, ΔT = T1 - T2 = 10°C - 3°C = 7°C, which needs to be converted to Kelvin (ΔT = 7K). We are also given the thermal conductivity (k) of the glass, 0.78 W/m·K and the time of heat loss (t) = 5 hours.
02

Use the formula for heat conduction

The formula for heat conduction is Q = (k * A * ΔT * t) / L, where Q is the heat loss through the glass. First, we need to convert L to meters: L = 0.5 cm = 0.005 m, and the time to seconds: t = 5 h * 3600 s/h = 18000 s.
03

Calculate the heat loss for 0.5 cm thick glass

Now plug in the given values to the formula: Q = (0.78 * 4 * 7 * 18000) / 0.005. Calculate the result: Q = 34128 Ws.
04

Repeat steps for 1 cm thick glass

For 1 cm thick glass, L = 1 cm = 0.01 m. Use the same formula: Q' = (0.78 * 4 * 7 * 18000) / 0.01. Calculate the new result: Q' = 17064 Ws.
05

Interpret the results

The heat loss through the 0.5 cm thick glass over a period of 5 hours is 34128 Ws, and the heat loss through the 1cm thick glass is 17064 Ws. The thicker the glass, the less heat loss occurs due to its increased resistance to heat conduction.

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Most popular questions from this chapter

Consider a 20-cm thick granite wall with a thermal conductivity of \(2.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature of the left surface is held constant at \(50^{\circ} \mathrm{C}\), whereas the right face is exposed to a flow of \(22^{\circ} \mathrm{C}\) air with a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\), and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Two surfaces, one highly polished and the other heavily oxidized, are found to be emitting the same amount of energy per unit area. The highly polished surface has an emissivity of \(0.1\) at \(1070^{\circ} \mathrm{C}\), while the emissivity of the heavily oxidized surface is \(0.78\). Determine the temperature of the heavily oxidized surface.

Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

While driving down a highway early in the evening, the air flow over an automobile establishes an overall heat transfer coefficient of \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The passenger cabin of this automobile exposes \(9 \mathrm{~m}^{2}\) of surface to the moving ambient air. On a day when the ambient temperature is \(33^{\circ} \mathrm{C}\), how much cooling must the air conditioning system supply to maintain a temperature of \(20^{\circ} \mathrm{C}\) in the passenger cabin? (a) \(670 \mathrm{~W}\) (b) \(1284 \mathrm{~W}\) (c) \(2106 \mathrm{~W}\) (d) \(2565 \mathrm{~W}\) (e) \(3210 \mathrm{~W}\)
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