Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 19, Problem 16SP

A certain double-pane window consists of two glass sheets, each 80  cm × 80  cm × 0.30  cm , separated by a 0.30-cm stagnant air space. The indoor surface temperature is 20 °C, while the outdoor surface temperature is exactly 0 °C. How much heat passes through the window each second? k T = 0.84  W/k m for glass and about 0.080 W/k m for air.

Expert Solution & Answer
Check Mark
To determine

The heat that passes through the double pan window in each second if it consists of two glass sheets, each 80 cm×80 cm×0.3 cm, separated by a 0.3cm stagnant of air space and the indoor space temperature is 20°C while the outdoor surface temperature is 0°C. Thermal conductivity of the glass plate is 0.84W/mK and thermal conductivity of air is 0.08 W/mK.

Answer to Problem 16SP

Solution:

72.5cal/s

Explanation of Solution

Given data:

The indoor surface temperature is 20°C.

The outdoor surface temperature is 0°C.

The dimension of the glass window is 80 cm×80 cm×0.30 cm.

The separation between the double pane window is 0.30 cm.

The thermal conductivity of the glass is 0.84W/mK.

The thermal conductivity of the air is 0.080W/mK.

Formula used:

Consider the expression for the equilibrium condition for heat flow from one plate to another:

kT1A1T1TL1=kT2A2TT2L2

Here, kT1 is the thermal conductivity of first plate, kT2 is the thermal conductivity of second plate, A1 is the area of first plate, A2 is the area of second plate, T1 is the temperature of first plate, T2 is the temperature of second plate, T is the temperature of the interface, L1 thickness of the first plate, and L2 thickness of the second plate.

Consider the expression for the heat passes through surface:

ΔQΔt=kT1A1T1TL1

Write the expression for the area of the rectangular ice sheet:

A=lb

Here, l is the length of the glass sheet and b is the width of the glass sheet.

Explanation:

Shape of the window is rectangular. Therefore, recall expression for the area of the window:

A=lb

Substitute 80cm for l and 80cm for b

A=(80 cm)(80 cm)=6400 cm2(104 m21 cm2)=0.64 m2

Consider the expression for the equilibrium condition for heat flow from air to glass plate:

kT1A1T1TL1=kT2A2TT2L2

Substitute 0.84W/mK for kT1, 0.08W/mK for kT2, 20°C for T1, 0°C for T2, 0.64m2 for A1, 0.64 m2 for A2, 0.3cm for L2, and 0.3cm for L1

(0.84 W/mK)(0.64m2)(20°CT0.30 cm(102 m1 cm))=0.08 W/mK(0.64m2)(T0°C0.3 cm(102 m1 cm))179.2(20T)=17.067T179.2T17.067T=3584196.267T=3584

Further solve for T

T=3584196.267=18.26°C=18.3°C

Temperature of the glass air interface is 18.3°C.

Consider the process of the heat transfer is steady. Therefore, the heat transfer rate will be constant.

Again, consider the expression of heat passes the window:

ΔQΔt=kT1A1T1TL1

Substitute 0.84W/mK for kT1, 20°C for T1, 0.64 m3 for A1, 0.3cm for L1, and 18.3°C for T .

ΔQΔt=(0.84 W/mK(1cal/scm°C418.4 W/mK))(6400 cm2)(20°C18.3°C0.3 cm)=72.5 cal/s

Conclusion:

Therefore, the amount of heat passes through the window is 72.5cal/s.

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