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
q in joules = mass air x specific heat air x (Tfinal-Tinitial)
Then 1 J = 2.78E-7 kWh.
To calculate the amount of energy required to heat the air in the room, we need to follow these steps:
Step 1: Calculate the change in temperature.
Temperature change (ΔT) = Final temperature - Initial temperature
ΔT = 28oC - 7oC
ΔT = 21oC
Step 2: Calculate the heat energy required.
Heat energy (Q) = Mass of air × Specific heat capacity × Temperature change
Q = 48 kg × 1.03 J/goC × 21oC
Step 3: Convert the heat energy to kilowatt-hours (KWh).
1 kilowatt-hour (KWh) = 3.6 × 10^6 joules (J)
Q (in KWh) = Q (in J) ÷ 3.6 × 10^6 J
Now, let's calculate the heat energy:
Q = 48 kg × 1.03 J/goC × 21oC
Q ≈ 1081.44 J
Q (in KWh) = 1081.44 J ÷ 3.6 × 10^6 J
Q (in KWh) ≈ 0.0003 KWh
Therefore, approximately 0.0003 KWh of energy are necessary to heat the air in the room from 7oC to 28oC.