Please show full working. Thanks The same scientist now wants to lift 1000 kg eq

Question

Please show full working. Thanks

The same scientist now wants to lift 1000 kg equipment to the top of the hill he is working on, a rise of 50 meters. He sets up a solar energy system with parabolic mirrors that collects sunlight and warms his working fluid to 120degree C (the working fluid docs not freeze and docs not boil in the temperature range over which it is being used). The exhaust temperature is - 20degree C. What is the minimum amount of energy required to heat his working fluid and run his engine to lift the mass?

Explanation / Answer

T1 (Temp of source) = 120+273 = 393 K

T2 (Temp of sink)= -20 + 273 = 253 K

Efficiency of heat engine = 1-(T2/T1)

= 1-(253/393)

=.35

Mass of equipment(M)= 1000 kg

Height of hill(h) = 50 m

Acc. due to gravity = 9.8 m/s2

Work done to lift equipment = m x g x h

= 1000 x 9.8 x 50

= 490000 J

Energy consumed in lifting = Work done in lifting/Efficiency of heat engine used

= 490000/.35

= 1400000 J

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