On a cold winter\'s day heat leaks slowly out of a house at the rate of 20.5 kW

Question

On a cold winter's day heat leaks slowly out of a house at the rate of 20.5 kW .

Part A

If the inside temperature is 21 C, and the outside temperature is -12.5 C, find the rate of entropy increase.

Express your answer using two significant figures.

SubmitMy AnswersGive Up

On a cold winter's day heat leaks slowly out of a house at the rate of 20.5 kW .

Part A

If the inside temperature is 21 C, and the outside temperature is -12.5 C, find the rate of entropy increase.

Express your answer using two significant figures.

SubmitMy AnswersGive Up

Explanation / Answer

Ti = 21 + 273.15 = 294.15

To = 273.15 - 12.5 = 260.65

d(Entropy) = dQ/T
So when, in 1 second, 20.5 kJ goes from outside to inside, the inside has lost 20.5 kJ, but the outside has gained 20.5 kJ.

Inside: dQ = -20.5/294.15
Outside: dQ = 20.5/260.65

Therefore, the net change in entropy in 1 second is:
20.5/260.65 - 20.5/294.15 = 0.008957 kJ/K
= 8.96 J/K

dE / dT = 8.96 W/K

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.