Physics, asked by rrgawali6p5ojcg, 1 year ago

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?

Answers

Answered by jahnvi111
0
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
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