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1). Compare the heat loss between a standard double pane window 0.9 m by 0.47 m hat has a gap filled with air that is 3 mm wide with a triple pane windows of the same size that has two gaps with air inside them, each 3 mm wide. The outside temperature is 8° C and the inside temperature is 25° C* for both situations. The glass between the two gaps is 4 mm wide.

1). Compare the heat loss between a standard double pane window 0.9 m by 0.47 m hat has a gap filled with air that is 3 mm wide with a triple pane windows of the same size that has two gaps with air inside them, each 3 mm wide. The outside temperature is 8° C and the inside temperature is 25° C* for both situations. The glass between the two gaps is 4 mm wide.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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1). Compare the heat loss between a standard double pane window 0.9 m by 0.47 m hat has a gap filled with air that is 3 mm wide with a triple pane windows of the same size that has two gaps with air inside them, each 3 mm wide. The outside temperature is 8° C and the inside temperature is 25° C* for both situations. The glass between the two gaps is 4 mm wide.
Transcribed Image Text:1). Compare the heat loss between a standard double pane window 0.9 m by 0.47 m hat has a gap filled with air that is 3 mm wide with a triple pane windows of the same size that has two gaps with air inside them, each 3 mm wide. The outside temperature is 8° C and the inside temperature is 25° C* for both situations. The glass between the two gaps is 4 mm wide.
Since heat flow is measured in watts (J/s), a simple model emerges if you throw in the obligatory substance-dependent constant k: kA(T2 – T1) P = d
Transcribed Image Text:Since heat flow is measured in watts (J/s), a simple model emerges if you throw in the obligatory substance-dependent constant k: kA(T2 – T1) P = d
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