Question

Take temperatures and efficiencies to be exact. In the winter, heat from a house with an inside temperature of 18 ∘C leaks out at a rate of 2.4×10^4 J/s . The outside temperature is 0∘C.

1. What is the change in entropy per second of the house?

2. What is the total change in entropy per second of the house-outside system?

Thank you!!

Answer 1

(a) When a system exchanges amount Q of heat at constant absolute temperature T, i.e. T in Kelvins, its entropy changes by:
∆S = Q/T
When you replace Q by the heat flow ∆Q/∆t you get a relation for the rate of change in entropy:
∆S/∆t = (∆Q/∆t) / T_house

The rate of change in entropy of the house is:
(∆S/∆t)_house = - |(∆Q/∆t)| / T_house

entropy change of house is negative as the heat is lost
= - 2.4×10⁴ Js⁻¹ / (18+ 273)K
= - 82.47 JK⁻¹s⁻¹

(b) The rate of change in entropy of the surrounding is:
(∆S/∆t)_surrounding = + |(∆Q/∆t)| / T_surrounding
= + 2.4×10⁴ Js⁻¹ / (0+ 273)K
= + 87.9JK⁻¹s⁻¹

The total change in entropy equals the sum of the changes of the partial systems:
(∆S/∆t)_total = (∆S/∆t)_house + (∆S/∆t)_surrounding
= - 82.47 JK⁻¹s⁻¹ + 87.9JK⁻¹s⁻¹
= 5.44JK⁻¹s⁻¹