(a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to
another environment at 45.0ºC ? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per 3.60×106J (a kilowatt-hour)?
a) First all the temperature units should be in Kelvin. -30°C + 273.15 = 243.15K and 45°C + 273.15 = 318.15K. Coefficient of performance for refrigerator is COP = TL/(TH-TL) where TL is Lower Temperature and TH is the Higher Temperature. COP = 243.15/(318.15-243.15) = 3.242
b) COP = Cooling Effect/Work Input. Cooling Effect (or Desired Output) is 4186 kJ.
So Work Input = Cooling Effect/COP = 4186KJ/3.242= 1291.178 KJ
c) 3.6 MJ (3600kJ) is for 10 cents, using ratio for cost: energy used
A ) t 1 = 2 h, t 2 = 6 h Δ t = t 2 - t 1 = 6 - 2 = 4 h 54 = 50 + a Δ t 54 = 50 + 4 a 4 a = 54 - 4 4 a = 4 a = 4 : 4 a = 1 km/h² v o = 48 km/h An equation that can be used to describe the velocity of the car at the different times is: v = 48 + t B ) The graph is in the attachment.
