Answer:
mass of water, mw = 300g = 0.3kg
∆Tw = (80 - 20) °C
volume of air, va = 100m³
mass of air, ma = 100g = 0.1kg
∆Ta = ?
H = mc∆T
Hw = mwcw∆Tw
Hw = 0.3*4200*60
Hw = 75600J
Hw = 75.6 kJ
All the above heat energy got absorbed by air,
that is; Ha = 75600J
since it's given that the heat was absorbed by a specific amount of volume of air
then specific capacity of volume of air is
then,
ca = Ha × density
ma temp
then,
Ha = vaca∆Ta
where, ca = volumetric heat capacity of air = 0.012kJ/m³°C
75.6kJ = 100m³ × 0.012kJ/m³°C × ∆Ta
75.6 = 1.2/°C × ∆Ta
∆Ta = 63°C
63°C is the temperature change in air.
