(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A system absorbs QH = 481 J of heat reversibly from a hot reservoir at temperature TH = 370 K, and gives off QC = 155.4 J of heat to a cold reservoir at TC = 156 K. During this process, W = 90.9 J of work is done by the system.

a) Find ΔIE, the change in internal energy of the system due to this process.

b) Find ΔSsystem, the change in entropy of the system due to this process.

c) Find ΔSuniverse, the change in entropy of the universe due to this process.

d) Suppose second system uses irreversible processes, but all the given values (for temperature and heat) are the same as for the original system. How would your answers to parts a, b and c change?

the answers to (a) and (c) are the same; the answer to (b) is increased

the answers to all questions are the same

the answer to (b) increases, the answers to (a) and (c) decrease

the answers to (b) and (c) are the same; the answer to (a) is decreased

the answers to all questions are increased

the answer to (c) increases, the answers to (a) and (b) decrease

the answers to (a) and (b) are the same; the answer to (c) is increased

the answers to all questions are decreased

the answers to (a) and (b) are the same; the answer to (c) is decreased

the answer to (c) decreases, the answers to (a) and (b) increase

the answer to (b) decreases, the answers to (a) and (c) increase

the answers to (a) and (c) are the same; the answer to (b) is decreased

the answers to (b) and (c) are the same; the answer to (a) is increased

the answer to (a) decreases, the answers to (b) and (c) increase

the answer to (a) increases, the answers to (b) and (c) decrease

2. Relevant equations

ΔI.E.=ΔQ+W

ΔS=ΔS_{h}+ΔS_{c}

3. The attempt at a solution

Using the first law of thermodynamic the heat added to the system would be 481J and the work done is -90.9j. i believe it is negative because it is work done by the system so pluging in values i get

ΔI.E.=481J - 90.9J = 390.1J

I thought this answer made sense because the system was gaining more heat than it was exhausting so the change in internal energy should be positive but webasign says its wrong i don't know what i am doing wrong. does it have something to do with the fact that it is reversible? should i be using carnot equations instead of F.L.T.? i got part b by adding the entropy of the hot and cold systems and got the correct answer and my notes say that for any reversable process it causes ΔS_{univ}=0 so i got part c as well part d i am thinking the cange in entropy of the universe will increase. I am only assuming this because of my ΔS equation stated above which would result in an answer greater than 0. i see no reason why an irriversable system would have a different internal energy and entropy than an equivilent reversable system but again i do no know because my equation for ΔI.E. is not working so maybe their is a different equation i need to be using. i got all my other H.W. answers on thermodynamics but this one is giving me great trouble it is probably something tiny i am not seeing like usual but as of right now i am totally lost any help would be greatly appreciated especial the week before finals Thank you in advance

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# Homework Help: Change in I.E. for a reversible process

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