# Refrigerator Question

xmonsterx

## Homework Statement

PROBLEM:
Refrigeration units can be rated in "tons". A 1-ton air conditioning system can remove sufficient energy to freeze 1 British ton (2000pounds) = 909 kg) of 0 degree C water into 0 degree C ice in one 24-h day.

QUESTION:
If, on a 40 degree C day, the interior of a house is maintained at 20 degree C by the continuous operation of a 5-ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs $0.10 per kWh and that the unit's coefficient of performance is 15% that of an ideal refrigerator. 1 kWh= 3.60 times 10^6 J. PLEASE GIVE THE ANSWER IN$/hr !

COP = Q_L / W

## The Attempt at a Solution

COP = (40 +273) / W ?

Homework Helper

## Homework Statement

PROBLEM:
Refrigeration units can be rated in "tons". A 1-ton air conditioning system can remove sufficient energy to freeze 1 British ton (2000pounds) = 909 kg) of 0 degree C water into 0 degree C ice in one 24-h day.

QUESTION:
If, on a 40 degree C day, the interior of a house is maintained at 20 degree C by the continuous operation of a 5-ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs $0.10 per kWh and that the unit's coefficient of performance is 15% that of an ideal refrigerator. 1 kWh= 3.60 times 10^6 J. PLEASE GIVE THE ANSWER IN$/hr !

COP = Q_L / W

## The Attempt at a Solution

COP = (40 +273) / W ?

First you have to find how much heat in joules that you have to remove in an hour. (Hint: how much heat does one have to remove from 909 kg of water at 0 C to turn it into ice at 0C?)

Then you have to determine the COP of the air conditioner to determine how much work must be performed to run it for one hour (in Joules and then KwH).

AM

xmonsterx
First you have to find how much heat in joules that you have to remove in an hour. (Hint: how much heat does one have to remove from 909 kg of water at 0 C to turn it into ice at 0C?)

Then you have to determine the COP of the air conditioner to determine how much work must be performed to run it for one hour (in Joules and then KwH).

AM

i don't know what formula to use to find how much heat is removed in an hour. I need additional help/ hints. thanks

Homework Helper
i don't know what formula to use to find how much heat is removed in an hour. I need additional help/ hints. thanks
You need to use the heat of fusion of water, which s 333.55 kJ/kg.

AM

xmonsterx
You need to use the heat of fusion of water, which s 333.55 kJ/kg.

AM

wait.. heat of fusion which is Lf = mL = 333.55 ... how does this help me solve for the answer?

what if i use deltaQ = mCdeltaT ?

delta Q = 909kg * 1cal/g C * ((40+273)-(20+273)) ??

does this look right? or am i totally off?

Homework Helper
wait.. heat of fusion which is Lf = mL = 333.55 ... how does this help me solve for the answer?

what if i use deltaQ = mCdeltaT ?

delta Q = 909kg * 1cal/g C * ((40+273)-(20+273)) ??

does this look right? or am i totally off?
You cannot use $\Delta Q = mC\Delta T$ because there is no change in temperature. Only a change of state. So you have to use:

$$\Delta Q = mL_f$$ where $L_f$ is the heat of fusion of water.

AM

xmonsterx
You cannot use $\Delta Q = mC\Delta T$ because there is no change in temperature. Only a change of state. So you have to use:

$$\Delta Q = mL_f$$ where $L_f$ is the heat of fusion of water.

AM

ok so...
delta Q = (909kg) * (333.55) = 3.0*10^5

so then do i plug this number in this equation --> COP= Q_L / W --> COP = Q_L / Q_H -Q_L ??

am i on the correct path??