(a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to
1 answer:
Answer:
a) 3.242
b) 1291.178 KJ
c) 3.59 cents
Explanation:
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
3.59 cents
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1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.
<u>Explanation</u>:
We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:
Better understood from numerical example as given:
If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?
This can be solved as follows:
It shows that man A will have more K.E.
Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.