Question

In a beaker, 300 grams of water at 80°C cools down to 20°C. Assume all the heat from the water is absorbed by 100 cubic meters of air, with a mass of 100.000 grams . What is the temperature change in the air?

Answer 1

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.