Question

A 4-ton air conditioner removes 5.06×10⁷ J (48,000 British thermal units) from a cold environment in 1.00 h. (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating (EER) of 12.0? (b) What is the cost of doing this if the work costs 10.0 cents per 3.60×10⁶ J (one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating (EER) of an air conditioner or refrigerator is defined to be the number of British thermal units of heat transfer from a cold environment per hour divided by the watts of power input.

a) 4.22×10⁶ J, $0.47
b) 4.94×10⁶ J, $0.55
c) 5.67×10⁶ J, $0.63
d) 6.39×10⁶ J, $0.71

Answer 1

To remove 5.06x10^7 J of heat from a cold environment in 1.00 h with an energy efficiency rating (EER) of 12.0, the necessary energy input in joules is 4.22x10^6 J. The cost of doing this is $0.47. This cost seems realistic.

the correct answer is option a) 4.22x10^6 J, $0.47.

Step-by-step explanation:

To calculate the energy input in joules necessary to remove 5.06x10^7 J of heat from a cold environment in 1.00 h, we can use the formula:

Energy input (J) = Heat transfer (J) / Energy efficiency rating (EER)

Using the given values, we can calculate:

Energy input (J) = 5.06x10^7 J / 12.0 = 4.22x10^6 J

To calculate the cost, we can use the formula:

Cost = Energy input (J) / 3.60x10^6 J (1 kilowatt-hour) * Cost per kilowatt-hour

Using the given values, we can calculate:

Cost = 4.22x10^6 J / 3.60x10^6 J * 0.10 dollars = 0.47 dollars

Based on these calculations, the correct answer is option a) 4.22x10^6 J, $0.47.