Verifying Efficiency in a Thermodynamic System: Q1 & Q2

[Falinox]

Homework Statement

http://i754.photobucket.com/albums/xx183/reddiesel08/Q2.jpg

Homework Equations

e = 1 - (Tc/Th)
(Temp in Kelvins)

The Attempt at a Solution

e = 1 - (300k/520k) = .42 or 42%
No because the engine can only perform 42 J of work for each 100 J of energy.

Homework Statement

http://i754.photobucket.com/albums/xx183/reddiesel08/Q1.jpg

Homework Equations

eff = 1 - (Tc/Th)
eff = W/Qh

The Attempt at a Solution

I know:
e = .35
Th = 180*C + Tc
Tc = Th - 180*

Need someone to verify number 1 for me and to help me solve number 2. Thanks in advance.

 

[LeonhardEuler]

Number 1 is right. For number 2, you have all the equations you need. You have
$$e = 0.35 = 1-\frac{T_C}{T_H}$$
and
Th = 180+ Tc
So you can solve them together by substitution.

 

[Falinox]

Number 1 is right. For number 2, you have all the equations you need. You have
$$e = 0.35 = 1-\frac{T_C}{T_H}$$
and
Th = 180+ Tc
So you can solve them together by substitution.

ok so...

.35 = 1 - [(Tc + 273)K / (Tc + 180 + 273)K]

.35 = 1 - [(Tc + 273K) / (Tc + 453K)]

-.65 = - [(Tc + 273K) / (Tc + 453K)] (divide by -1 on both sides)

.65 = [(Tc + 273K) / (Tc + 453K)]

.65(Tc + 453K) = Tc + 273K

.65Tc + 294.45K = Tc + 273K

21.45K = .35Tc

Tc = 61.29K (convert to *C) = -211*C

This can't be right. Any suggestions on what I'm doing wrong?

 

[LeonhardEuler]

You're units are not right. If you put in the "+273", then you are already solving for the temperatures in Celsius, so you don't need to convert in the end. Alternatively, and I think this is easier, you could solve it without the "+273"'s to get the answer in Kelvin, and convert in the end. But you got the right answer in Celsius before you converted.