To help keep his barn warm on cold days, a farmer stores 802 kg of solar-heated water (Lf = 3.35E+5 J/kg) in barrels. For how many hours would a 2.19 kW electric space heater have to operate to provide the same amount of heat as the water does, when it cools from 13.7 to 00C and completely freezes?
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In this case, the energy produce by the electric space heater must be the same with the energy the water release.
Energy released when water cools down from 10 to 0 degree.
Q1 = mcdT = 865 x 4200 x 10 = approx 3.6E+7 Joules
Energy released when water freezes
Q2 = mLf = 865 x 3.35E+5 = approx 2.9E+8 Joules
Total = Q1 +Q2 = 3.26E+8
This amount of energy is the energy that the space heater needs to produce
3.26E+8 = Power x time
time = 174331.55 seconds
time = 48.4 hours
Energy released when water cools down from 10 to 0 degree.
Q1 = mcdT = 865 x 4200 x 10 = approx 3.6E+7 Joules
Energy released when water freezes
Q2 = mLf = 865 x 3.35E+5 = approx 2.9E+8 Joules
Total = Q1 +Q2 = 3.26E+8
This amount of energy is the energy that the space heater needs to produce
3.26E+8 = Power x time
time = 174331.55 seconds
time = 48.4 hours
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