Explanation Given info: The average thermal conductivity of the walls and roof of the house is 0.480 W / m ⋅ ° C , the thickness of the walls and the roof of the house is 21.0 cm , the heat of combustion is 9300 kcal / m 3 , the length of the wall is 10.0 m , the height of the walls is 5.00 m , the value of the breadth is 8.00 m , the angle of the roof is 37.0 ° , the inside temperature is 25.0 ° C and the outside temperature is 0.0 ° C . Write the expression for the area of the front wall. A 1 = b × h Here, b is the breadth. h is the height. Substitute 8.00 m for b and 5.00 m for h in above equation. A 1 = 8.00 m × 5.00 m = 40.0 m 2 Thus, the area of the front wall is 40.0 m 2 . Write the expression for the area of the side wall. A 1 = l × h Here, l is the length. h is the height. Substitute 10.00 m for l and 5.00 m for h in above equation. A 2 = 10.00 m × 5.00 m = 50.0 m 2 Thus, the area of the side wall is 50.0 m 2 . Write the expression for the area of the roof. A 3 = l × ( b 2 cos 37 ° ) Substitute 10.00 m for l and 8.00 m for b in above equation. A 3 = 10.00 m × ( 8.00 m 2 cos 37 ° ) ≈ 50.0 m 2 Thus, the area of the roof is 50.0 m 2 . Write the expression for the area of the triangle shape of the wall. A 4 = 1 2 [ b × b 2 × tan 37 ° ] Substitute 8.00 m for b in above equation. A 4 = 1 2 [ 8.00 m × 8.00 m 2 × tan 37 ° ] ≈ 12.0 m 2 Thus, the area of the triangle shape of the wall is 12.0 m 2 . Write the expression for the total area of the home. A = 2 ( A 1 + A 2 + A 3 + A 4 ) Substitute 40.0 m 2 for A 1 , 50.0 m 2 for A 2 , 50.0 m 2 for A 3 and 12

9th Edition

ISBN: 9780100454897

Author: Jewett

Publisher: YUZU

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Chapter 20, Problem 20.78AP

To determine

The volume of gas which is burned each day.

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Chapter 20 Solutions

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The volume of gas which is burned each day.

Solution Summary: The author explains the average thermal conductivity of the walls and roof of a house and the heat of combustion.

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The volume of gas which is burned each day.

Chapter 20 Solutions