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?

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.

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.