Question

# Thermal Storage Solar heating of a house is much more efficient if there is a way...

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 solar-heated 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?

 7200 kg

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?

57.1857.18tt

Incorrect; Try Again; 3 attempts remaining

Part C

Model the concrete slab as being surrounded on both sides (contact area 24 m2) with a 2.4-m-thick 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?

nothingnothing

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?

2.92.9wattwatt

Incorrect; Try Again; 4 attempts remaining

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

=0.02*24*5/2.4=1W

sigma=stefan constant=5.67 x 10 -8

=0.91*5.67 x 10 -8 *24*(57.18^4-52.18^4)

=4.1W

#### Earn Coins

Coins can be redeemed for fabulous gifts.