Thermal Storage Solar heating of a house is much more efficient if there is a way to store the thermal energy collected during the day to warm the house at night. Suppose one solarheated home utilizes a concrete slab of area 12 m2 and 25 cm thick.
Part A
Part complete
If the density of concrete is 2400 kg/m3, what is the mass of the slab?
Express your answer to two significant figures and include appropriate units.
7200 kg 
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Correct
Part B
The slab is exposed to sunlight and absorbs energy at a rate of 1.9 ×107J/h for 10 h. If it begins the day at 22 ∘C and has a specific heat of 750 J/(kg⋅K), what is its temperature at sunset?
Express your answer to two significant figures and include appropriate units.


57.1857.18tt 
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Part C
Model the concrete slab as being surrounded on both sides (contact area 24 m2) with a 2.4mthick layer of air in contact with a surface that is 5.0 ∘C cooler than the concrete. At sunset, what is the rate at which the concrete loses thermal energy by conduction through the air layer?
Express your answer to two significant figures and include appropriate units.


nothingnothing 
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Part D
Model the concrete slab as having a surface area of 24 m2 and surrounded by an environment 5.0 ∘C cooler than the concrete. If its emissivity is 0.91, what is the rate at which the concrete loses thermal energy by radiation at sunset?
Express your answer to two significant figures and include appropriate units.


2.92.9wattwatt 
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Part B
1.9*10^7J/h for 10h
Total heat=1.9*10^8J
1.9*10^8=mc(delta t)
1.9*10^8=7200*750*(delta t)
(delta t)=35.19
Final temperature=22+35.19
=57.185degree celsius
=57degree celsius(2 significant digits)
Part C
Heat loss through air layer by conduction=KAdt/dx
K=conductivity of air=0.02W/mK
A=area
dt/dx=temperature gradient
=0.02*24*5/2.4=1W
D)Heat loss by radiation=emissivity(sigma)(Area)((T_{object})^4T_{surrounding})^4
sigma=stefan constant=5.67 x 10 ^{}^{8}
=0.91*5.67 x 10 ^{}^{8} *24*(57.18^452.18^4)
=4.1W
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