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For a certain ideal Carnot engine, the hot reservoir is 35°C higher than the cold reservoir. If this engine is to have an efficiency of 20%, what must be the temperature of the hot reservoir?

Respuesta :

obasola1

Answer:

Temperature of the hot reservoir is 1540K

Explanation:

[tex]E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\

Efficiency of a carnot engine is given by the above

Tc=temperature of the cold reservoir

Th= temperature of the hot reservoir

K=273+ 35  (convert  35°C to kelvin)

K=308k

{T_h}={T_c}+308-----------------------(equation  1)

20%=1-{T_c}/{T_h}[/tex]

0.2=({T_c}+308-{T_c})/{T_c}+308

.2({T_c}+61.6=308

0.2{T_c}=246.4

{T_c}=1232

recall from equation 1

{T_h}=308+1232

{T_h}=1540K

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