How Efficient Can a Heat Engine Be Between 1200 K and 300 K?

[science.girl]

Homework Statement

A heat engine is built to operate between temperatures of 1200 K and 300 K. It is used to lift a 30 kg mass at a constant velocity of 4 m/s.

a) Determine the power that the heat engine must supply to lift the mass.

b) Determine the maximum possible efficiency of the heat engine.

c) If the engine were to operate at the maximum possible efficiency, determine the following:
i. The rate at which the hot reservoir supplies heat to the engine.
ii. The rate at which heat is exhausted to the cold reservoir.

Homework Equations

P = W/t
e = 1 - T₁/T₂

The Attempt at a Solution

I figured out that (b) is 1 - (300/1200) = 75%.

For (a), there is no distance stated to figure out work (W = Fd), and there is no time to determine power (P = W/t). What am I missing?

 

[rl.bhat]

Power= rate of doing work. = Fd/t = fv.

 

[science.girl]

Power= rate of doing work. = Fd/t = fv.

Oh! So, P = Fv = (30 kg)(9.8 m/s^2)(4 m/s) = 1176 J/s.

So, if the power supplied to lift the mass is 1176 J/s, and the maximum possible efficiency is 75%, then the rate heat is removed from the hot reservoir... Wouldn't that depend on the first law of thermodynamics?
[delta]U = Q + W

But how would you apply the equation exactly, and what would it signify?

 

[Andrew Mason]

Oh! So, P = Fv = (30 kg)(9.8 m/s^2)(4 m/s) = 1176 J/s.

So, if the power supplied to lift the mass is 1176 J/s, and the maximum possible efficiency is 75%, then the rate heat is removed from the hot reservoir... Wouldn't that depend on the first law of thermodynamics?
[delta]U = Q + W

But how would you apply the equation exactly, and what would it signify?

You do not need to apply the first law here. You just have to apply the efficiency. The work output is 1176 watts (J/s) at 75% efficiency (efficiency = work out/energy in) so the energy (heat flow) input must be...?

AM

 

[science.girl]

You do not need to apply the first law here. You just have to apply the efficiency. The work output is 1176 watts (J/s) at 75% efficiency (efficiency = work out/energy in) so the energy (heat flow) input must be...?

AM

So, given the equation e = W/Q_H...

.75 = (1176W)/Q_H

Q_H = 1568 WAnd, for the rate at which heat is exhausted to the cold reservoir, does this also depend on an efficiency equation?

 

[djeitnstine]

75% = 75/100 = 0.75

for rate at which heat is exhausted note that W = QH-QC. So dW/dt = dQH/dt - dQC/dt

 

[science.girl]

75% = 75/100 = 0.75

for rate at which heat is exhausted note that W = QH-QC. So dW/dt = dQH/dt - dQC/dt

Hmm... Unfortunately, this is a non-calc based course, so would you know of another approach?

 

[djeitnstine]

dW/dt = Power

dQh/dt = rate of energy output from the hot resevior

or perhaps delta Qh/ delta t which is average energy over a period of time

similar argument for Qc

 

[science.girl]

dW/dt = Power

dQh/dt = rate of energy output from the hot resevior

or perhaps delta Qh/ delta t which is average energy over a period of time

similar argument for Qc

Yes, but unfortunately, this is an algebra-based course. Is there another method for finding the rate at which heat is exhausted to the cold reservoir?

 

[Andrew Mason]

Yes, but unfortunately, this is an algebra-based course. Is there another method for finding the rate at which heat is exhausted to the cold reservoir?

W = Qh-Qc

P = W/t = Qh/t - Qc/t

You know two of those terms. So find Qc/t

AM

 

[science.girl]

W = Qh-Qc

P = W/t = Qh/t - Qc/t

You know two of those terms. So find Qc/t

AM

W = Q_H - Q_C

392 W = 980 W - Q_C/t

Q_C/t = 588 W

Correct?

 

[Andrew Mason]

W = Q_H - Q_C

392 W = 980 W - Q_C/t

Q_C/t = 588 W

Correct?

I am not sure what you are doing here.

W/t = 1176 w
Qh/t = W/e = 1568 w

AM

 

[science.girl]

I am not sure what you are doing here.

W/t = 1176 w
Qh/t = W/e = 1568 w

AM

My apologies. I was working on two similar problems... that's why the values were different.

Thank you very much for your help! I appreciate it.

-Science.girl