Question

A Carnot engine has a power output of 150 kW. The engine operates between two reservoirs at 20.0°C and 500°
c. (
a. How much energy does it take in per hour

Answer 1

Efficiency η of a Carnot engine is defined to be: 
η = 1 - Tc / Th = (Th - Tc) / Th 
where 
Tc is the absolute temperature of the cold reservoir, and 
Th is the absolute temperature of the hot reservoir. 

In this case, given is η=22% and Th - Tc = 75K 
Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). 

Th = (Th - Tc) / η 
Th = 75 / 0.22 = 341 K (rounded to closest number) 
Tc = Th - 75 = 266 K 

Lower temperature is Tc = 266 K 
Higher temperature is Th = 341 K

Answer 2

Answer:

8.8 × 10³³J

Explanation:

efficiency of an engine η  = net work output ÷ heat input = 1 - $\frac{T_{c} }{T_{H} }$

$T_{H}$ = 500 + 273 =773K

$T_{C}$ = 20 + 273 =293K

$n_{c}$ = 1 - $\frac{293}{773}$

  =0.62

power output= work output = 150kW

power output per hour = 150 × 1000×3600

                                     =5.4×10⁸J

recall, net work output ÷ heat input = 1 - $\frac{T_{c} }{T_{H} }$

∴ 0.62= 5.4×10⁸ ÷ heat input

heat input= 5.4×10⁸÷0.62

=8.81×10⁸J