How many bricks must be incorporated into the interior of the house? - passive solar house design
A house has been designed to have properties of the passive use of solar energy. The work will be inside the house included to act as a heat absorber. Each brick weighs approximately 1.8 kilograms. The specific heat of the stone is 0.85 J / g ⋅ K: How many stones are to be included in the interior of the house to offer the same thermal performance of a total of 1.4 x10 ^ 3 liters of water?
Tuesday, January 12, 2010
Passive Solar House Design How Many Bricks Must Be Incorporated Into The Interior Of The House?
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1,4 x10 ^ 3 liter water@3.784 liters / gallon = 5.3 x 10 ^ 3 liter
= 5.3 x10 ^ 6 ml = 5.3 g x 10 ^ 6
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5.3 x 10 ^ 6 g) (4.18joules / GC) = 2.21 x 10 ^ 7 J / C
Water
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Bricks:
(1800 g) (0.85 J / gK) = 1530 joules per brick
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1,4 x10 ^ 3 liter water@3.784 liters / gallon = 5.3 x 10 ^ 3 liter
= 5.3 x10 ^ 6 ml = 5.3 g x 10 ^ 6
----------------------
5.3 x 10 ^ 6 g) (4.18joules / GC) = 2.21 x 10 ^ 7 J / C
Water
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Bricks:
(1800 g) (0.85 J / gK) = 1530 joules per brick
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