Change in I.E. for a reversible process

[gravrider]

Homework Statement

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

Homework Equations

ΔI.E.=ΔQ+W
ΔS=ΔS_h+ΔS_c

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

 

[ehild]

The system gives off heat to the cold reservoir. Haven't forgotten about it?

ehild

 

[gravrider]

So would it be Q_h - Q_c + Work = ΔI.E.

 

[ehild]

Yes, but the work is negative...

ehild

 

[gravrider]

yea that is what i meant about work. thank you so much for your help hearing the solution now i feel so dumb for not seeing that earlier because the change in internal energy is equal to ΔQ not Just the heat entering the system and then if the process was irreversible the change in entropy of the universe would no longer be zero it would be .304 right so it would be greater and the other two would stay the same because the temperature and the heat are still the same right?