Question

a refrigerator has to transfer an average of 263 j of heat per secound from -10 to 25 centgrade.
calculate the average power consumed. asumed that the refrigerator is ideal
a.25w
b.30w
c.35w
d.40w
e.45w
@AllTehMaffs

Answer 1

that's the question @sarah786 posted, right? I can't figure out how to do this one for some reason.

Answer 2

yeah @AllTehMaffs it was in my test I was unable to solve it

Answer 3

Do you have any idea @Lessis ?

Answer 4

It has something to do with reversed Carnot cycles

First law of thermo is
\[Q_h=Q_c+W\]

and we want
\[P_{avg} = \frac{W}{t}\]

But it gives you a power in the first place, right? 263J/s

Just can't figure out how to relate that to anything :/

Answer 5

@AllTehMaffs it's tough

Answer 6

it's driving me crazy 0_o

Answer 7

I'm not sure about what you're supossed to do with the given temperatures.

Also, an ideal refrigerator is 100% efficient. Right?

Answer 8

yeah
\[COP = 1 = \frac{T_c}{T_h-T_c}=\frac{Q_c}{W}\]

Answer 9

\[Efficiency = \frac{ Q_{1} }{ Q_{2}-Q_{1} }=\frac{ Q_{1} }{ W } \]

Answer 10

for a fridge, yah

Answer 11

\[ \frac{Q_h}{Q_c} \propto \frac{T_h}{T_c}\]

Answer 12

\[\frac{1}{\frac{Q_h}{Q_c} - 1} \propto \frac{1}{\frac{T_h}{T_c}-1} = \frac{T_c}{T_h-T_c}\]

Answer 13

http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=th&chap_sec=05.5&page=theory

Answer 14

err,
\[ \frac{Q_h}{Q_c}≥\frac{T_h}{T_c}\]

Answer 15

Second law of thermo - entropy can increase but not decrease, so in a heat engine, the entropy it expels has to be at least a much as the entropy it absorbs

entropy =

\[S=\frac{Q}{T}\]

\[S_c=\frac{Q_c}{T_c}
\\ \
\\ S_h=\frac{Q_h}{T_h}\]
\[S_c≥S_h\]

\[\frac{Q_c}{T_c}≥\frac{Q_h}{T_h}\]

\[\frac{Q_c}{Q_h}≥\frac{T_c}{T_h}\]

Answer 16

but still dunno what to do with that freaking 263J/s :P

Answer 17

Huh. Well, the answer is 35 W. I guess. xD

Answer 18

I THINK THIS IS WHAT YOU HAVE TO DO.

\[COP = \frac{ T_{c} }{ T_{h} - T_{c} } \approx 7.51\]
\[COP = \frac{ Q_{c} }{ W }\]
\[\frac{ Q_{c} }{ W } = \frac{ Q_{c}/t }{ P } = COP\]
\[P = \frac{ Q_{c}/t }{ COP } \approx 35 W\]

Answer 19

WINNER!!!

Answer 20

Oh. And Qc/t is the head rejected per second, which is 263 J/s.

Answer 21

ooooohhhh

Answer 22

@sarah786 an answer!!

Answer 23

I really just divided 263/7.5 to see what happened though. xD
And then I just deduced the equation from there.

Answer 24

guess and check is the best way to math ^_^

Answer 25

Good thing this was a multiple choice question, huh? D:

Answer 26

For reals - the one benefit of them :P

Answer 27

Thank you so much @Lessis @AllTehMaffs

Answer 28

@AllTehMaffs @Lessis thanks :P