# Quick check on equations

1. Aug 25, 2008

### annjolino

1. The problem statement, all variables and given/known data

ideal heat engine takes in heat Qin at a temp Th. It exhausts heat Qout
a) how much work is done by the engine
b) What is the effiency of the engine
c) what is the exhaust temp of the engine

2. Relevant equations
W = Qin - Qout
Effiency = Qout/Qin
E = Th-Tc/Th

3. The attempt at a solution

i am pretty sure the frst two equations should give me what i need however its the third one that its a little stuck

I need to find Exhaust Temp which i think shoud be Tc...
so i rearranged the equation as shown

E=Th-Tc/Th
E x Th=Th-Tc
E x 2Th = -Tc
Tc= E x -2Th

when i factor in the values Qin = 460J Qout = 285J and Th 600K
i get a very cold answer and a number that shouldnt exist as a temp of -743.5K

where have i gone wrong???

2. Aug 25, 2008

### jianxu

Taking a quick glance at your math:

E x 2Th = -Tc <-----you made a mistake in this mathematical step. the Th on the right side can only subtract. therefore cannot be E * 2Th. Hope that helps ^^

3. Aug 25, 2008

### annjolino

so then would it be closer to

Tc= E x Th - Th

4. Aug 25, 2008

### jianxu

yes, remember there is still a negative sign on Tc and if you like, you can also factor the Th but is not necessary(mostly just to make it easier to read). Give that a try and see how it turns out =P

5. Aug 25, 2008

### annjolino

ok now when i put the values in the new equation
-Tc= E x Th - Th
-Tc= (0.62 x 600K) - 600K
-Tc = -228K

could i then make them both positive as they would cancel each other out??
Tc= 228K

6. Aug 25, 2008

### jianxu

yep ^_^

7. Aug 25, 2008