Question

a) What is the best possible coefficient of performance for a refrigerator that cools an environment at -28 degrees Celsius and exhauts heat to another environment at 37 degrees Celsius?

b) How much work in Joules must be done for a heat transfer of 4.925x10^6 J from the cold environment?

c) What is the cost of doing this, if the work costs 16.5 cents per 3.60x10^6 J (a kilowatt-hour)?

d) How many Joules of heat transfer occurs into the warm environment?

Answer 1

T₂= 245K

T₁= 310 K

COEFFICIENT OF PERFORMANCE = T₂/T₁- T₂

                                                                                = 245/310-245

                                                            = 3.76

COEFFICIENT OF PERFORMANCE= HEAT AT LOW TEMP/WORK DONE

3.76 = 4.925 X 10⁶/WORK

WORK DONE = 1.30 X10⁶ J

COST OF DOING = 1.3 X 10⁶/ 3.6 X 10⁶ *16.5

                                    = 5.98 CENTS

Coefficient of performance = heat gained by cold body/heat lost by hot body