How long will it take to heat a room from 10 0C to 20 0C if the heater consumes 12.5 X 105 J?
Specific heat capacity of air is 1.006 kJ / kg 0C
To calculate mass of air:
density of air is 0.036 kg /
ft3
Let the total volume of the room is 10 x 10 x 10 = 1000 ft3
Thus mass = density x volume
= 0.037 kg / ft3x 1000 ft3 = 37 kg
Temperature difference = 10 0C (20 0C - 10 0C)
Energy needed, Q = mass x specific heat capacity of air x change in temperature
Q = 37 kg x 1.006 kJ / kg 0C x 10 0C
= 372.22 kJ
Power is energy transferred / second.
Assuming, the heater consumes 12.5 x 105 J of heat in one hour, then time required to heat the room is calculated as follows.
12.5 x 105 J --> 3600 second
372.22 kJ ---> a
thus a = 0.2977 hour.
d. How many kilowatt hours of electricity is used during 0.2977 hour?
Thus Electrical energy = power x time
= 0.2722 kW x 0.2977 hour
= 0.081 kWh
With 25 cents per kWh, the cost of running the heater is = 0.081 kWh x 25 cents / kWh
= 2.025 cents
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