Question

The air in a kitchen has a mass of 60.0kg and a specific heat of 1505J/(kg°C).

What is the change in thermal energy of the air when it warms from 24°C to 31°C?



Question 10 options:

30,000J


100,835J


632,100J


90,300J

Answer 1

The formula to calculate the change in thermal energy is:

ΔE = mcΔT

where ΔE is change in thermal energy, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Here,
m = 60.0 kg
c = 1505 J/(kg°C)
ΔT = (31°C - 24°C) = 7°C

ΔE = (60.0 kg) * (1505 J/(kg°C)) * (7°C) = 632,100 J

Therefore, the change in thermal energy of the air when it warms from 24°C to 31°C is 632,100J.

Hence, the correct option is (C) 632,100J.

(If this doesn’t seem right to you comment!)

Answer 2

Answer: D

Explanation:

So you have 60kg of air, and the specific heat per kilogram is 1505j/(kg°C), this means you need 1505j to increase 1°C to the temperature of a mass of 1kg of air.

So, if the change in temperature is from 24°C to 31°C, you are increasing 7°C to 60kg of air.

then, for each degree you need 60*1505 j= 90300j

and for 7°c you need 7*90300j = 632100j, wich is answer D.