## 9.101 sign?

$\Delta S = \frac{q_{rev}}{T}$

Wilson Yeh 1L
Posts: 42
Joined: Fri Sep 29, 2017 7:06 am

### 9.101 sign?

A heat pump heats a house in winter by extracting heat from the cold outdoors and releasing it into the warm interior. For the transfer of a given amount of heat, (a) how do the entropies of the interior and exterior of the house change (increase or decrease)? (b) Which change is greater? Assume that the temperatures inside and outside the house do not change. Explain your answers.

In order to solve part b I know you're set it up as T(int) = q/S(int) and T(ext) = q/S(ext). However, why isn't the sign for q different for exterior and interior different? Why isn't it T(ext) = -q/S(ext)?

Janine Chan 2K
Posts: 71
Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.101 sign?

I believe the signs don't change due to the way the question is worded. We don't assign the interior or the exterior to be the "system," so we don't say the ∆S of the "surroundings" is -∆S of the "system." The question simply asks which is greater: the change in entropy of the exterior or the change in entropy of the interior. We evaluate the relative values of ∆S exterior and the ∆S interior by considering how high the temperature is in each region.