Entropy Change in Heat Pump: House Heating and Exterior Effects

AI Thread Summary
In a heat pump system, heat is extracted from the cold outdoors and transferred indoors, leading to a positive change in entropy for the interior. The exterior experiences a greater increase in entropy despite the heat being transferred, due to the lower temperature of the outdoor environment. This is explained by the relationship between temperature and entropy change, where a larger temperature difference results in a greater entropy change for the colder area. The confusion arises from the assumption that heat loss from outside equates to heat gain inside, but the temperature differential plays a crucial role. Thus, the entropy change is indeed greater outside the house.
3ephemeralwnd
Messages
27
Reaction score
0

Homework Statement



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.

Homework Equations



delta S = detal H / T

The Attempt at a Solution



since change in entropy is proportional to chnange in enthalpy, if the house is being heated, there is a positive change in enthalpy, and therefore an increase in entropy

but for part b, the answer in the book says change in entropy is greater OUTside of the house. why is that? the heat lost from outside is the heat gained inside, so shouldn't this be the case for entropy as well?
 
Physics news on Phys.org
3ephemeralwnd said:
but for part b, the answer in the book says change in entropy is greater OUTside of the house. why is that? the heat lost from outside is the heat gained inside, so shouldn't this be the case for entropy as well?

Not if the temperatures are different.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top