Question #55642

A room contains 48 kg of air. How many KWh of energy are necessary to heat the air in the house from 7oC to 28oC? The heat capacity of air is 1.03 J/goC.

Expert's answer

Q = C*m*(T2 - T1) = 1.03 J/(g*K)*48 kg*(28 C - 7 C) = 1.03 J/(g*K)*48000 g*(28 C - 7 C) = 1038240 J = 288.4 Wh = 0.2884 kWh

1 Wh = 3600 J

1 Wh = 3600 J

## Comments

Assignment Expert09.11.17, 10:33Dear visitor,

please use panel for submitting new questions

Maegan09.11.17, 04:02At what velocity (m/s) must a 28.7 kg object be moving in order to possess a kinetic energy of 1.03 J?

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