Minimum consumption rate of a heat pump

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Effiency of a heat pump is given as $\eta = \frac { T_h }{T_h – T_c} = 30$

W * 30/s = 6000*4.2 cal /s

Minimum consumption rate = W/s = 840 watt

Is this correct?

 

[rude man]

Homework Statement

Effiency of a heat pump is given as $\eta = \frac { T_h }{T_h – T_c} = 30$

This formula is not quite correct. It results in only about a 3% error in this case but it could be a lot more if T_h - T_c were a lot larger.

So actually none of the choices is particularly good. But maybe you picked the closest ...

 

[Pushoam]

This formula is not quite correct.

The question is asking for minimum consumption rate.
For this I have to use a heat pump with carnot engine.
For this case, isn't the followiing right ?

Effiency of a heat pump is given as $\eta = \frac { T_h }{T_h – T_c} = 30$

The textbook by blundell and blundell says,

 

[rude man]

OK I see the problem I think.
Your textbook defines efficiency as e = Q_h/W = (Q_c + W)/W for some reason.
But most textbooks talk about the "coefficient of performance" (COP) which is much more logical since this is heat removed per unit of work.
So per your textbook e = 30 but for the COP it's 29.
The word "efficiency" really doesn't apply here IMO since by definition efficiency can never exceed unity.

 

[Pushoam]

What is meant by IMO?

 

[rude man]

What is meant by IMO?

" In My Opinion".
Many such abbreviations were started with texting. This one is extremely widely used.