How to convert power to energy in a heat pump problem?

[Tangeton]

Homework Statement

A heat pump fitted with a 120W electric motor supplies 600W of heat per second to a room from outdoors. Calculate: a) the coefficient of performance of the heat pump
b) the heat gained from outdoors.

Homework Equations

Coeff of performance for heat pump = Qin/ W = Qin/(Qin-Qout)
Qin = Qout + W

The Attempt at a Solution

(a)
P = 600J/s
W = 120

P = Wt --> 600 = 120t --> t = 0.2s
Qout = Pt = 600 x 0.2 = 120J

Qin = 120 + 120 = 240

Qin/Qin - Qout = 240/240-120 = 2

The answer is 5.

(b)
120.

The answer is 480W
Please help...

 

[BvU]

Dear T,

Unfortunately, the errors start already with the rendering of the problem statement: you can't "supply 600W of heat per second" :
either you supply 600 Watt or you supply 600 J/s. No wonder, because $1W \equiv 1J/s$.

My guess: you misread or mistyped.

It gets a little worse when you write "P = Wt --> 600 = 120t --> t = 0.2s" . This is the wrong calculation, but on top of that it's executed wrong. You should be able to check the results from your calculator, e.g. 600 divided by approximately 100 should be approximately 6. (Or you know that 5 x 12 = 60 so 5 x 120 = 600).

And -- considering the above -- including the dimensions: 600 W / 120 W = 5 (without a dimension: it's a ratio).

Funny that this 5 looks a lot like the answer to (a). Any idea why ?

(b) is easy now, isn't it ?

 

[Tangeton]

Dear T,

Unfortunately, the errors start already with the rendering of the problem statement: you can't "supply 600W of heat per second" :
either you supply 600 Watt or you supply 600 J/s. No wonder, because $1W \equiv 1J/s$.

My guess: you misread or mistyped.

It gets a little worse when you write "P = Wt --> 600 = 120t --> t = 0.2s" . This is the wrong calculation, but on top of that it's executed wrong. You should be able to check the results from your calculator, e.g. 600 divided by approximately 100 should be approximately 6. (Or you know that 5 x 12 = 60 so 5 x 120 = 600).

And -- considering the above -- including the dimensions: 600 W / 120 W = 5 (without a dimension: it's a ratio).

Funny that this 5 looks a lot like the answer to (a). Any idea why ?

(b) is easy now, isn't it ?

So 600 is the Qin, and work done 120?
I want to use the equation somehow but I just literally cannot understand what part of the equation those numbers are...

 

[Tangeton]

You know what, its funny because I mistakenly said its not 600W but 900W in my first tries, which is why I failed to get the right answer...

 

[BvU]

I take it this exercise was given to you in the context of a lecture or a chapter on heat pumps or so ?

For a heat pump, in $CoP = {Q\over W}$ Q is the heat delivered and W is the work consumed by the heat pump.

It's kind of the opposite of a Carnot engine, where Qin is picked up, Qout is given off and W is delivered.

Carnot efficiency can never be > 1 , but a CoP can.

its funny because I mistakenly said its not 600W but 900W in my first tries, which is why I failed to get the right answer...

Work a bit slower, check all and every dimensions; before using a calculator, estimate the outcome of calculations. Try not to work too fast.