# Find amount of work necessary to extra 4000J from a body

1. Apr 8, 2010

### mahdert

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

Find the minimum amount of work required to extract 4000 J of heat from a body at 0 deg F, when the temprature of the enviroment is 100 deg F.

2. Relevant equations

W = (1 - t2/t1) * Q1

3. The attempt at a solution
instructor says 870.19, how??

Last edited: Apr 8, 2010
2. Apr 8, 2010

### Mapes

Re: Thermo

Hi mahdert, welcome to PF. According to your equation, it should take the same amount of work to transfer heat between reservoirs at 0K and 100K as it would for reservoirs at 0K and 200K. So right away it's clear that this is the wrong equation; you can't get transfer any hear from a reservoir at 0K.

Also, note the units.

3. Apr 8, 2010

### mahdert

Re: Thermo

The unit for the temperatures, you mean. I expressed them in Farenheits (deg F)..That will be t1 = 310.92 K and t2=255.37 K. Since heat is being drawn from a body with a lower temperature than the resevour, we need to supply work. The minimum amount of work would be what we would need for a carnot cycle. And I believe that is the correct formula. The answer I got now is 795 J.

Last edited: Apr 8, 2010
4. Apr 8, 2010

### Mapes

Re: Thermo

$0^\circ\mathrm{F}\neq 273\,\mathrm{K}$, and it's not the correct formula. Think of it this way: you're removing $Q$ amount of energy and $Q/T_2$ amount of entropy from the cold reservoir, and all the entropy goes into the hot reservoir: $(Q+W)/T_1$. At best (maximum efficiency in a Carnot cycle), these entropy amounts are equal. Solve for $W$.

EDIT: OK, I see you caught the temperature error.

5. Apr 8, 2010

### mahdert

Re: Thermo

thanks for the explanation.. i understand now..

6. Apr 8, 2010

Re: Thermo

Great!