Question

Steam locomotives have an efficiency of 17.0% and operate with a hot steam temperature of 425ºC.

(a) What would the cold reservoir temperature be if this were a Carnot engine?
(b) What would the maximum efficiency of this steam engine be if its cold reservoir temperature were 150ºC?

Answer 1

a) 570.46 deg K or 297.31 deg C

b) 0.393 or 39.3%

Step-by-step explanation:

A) Efficiency=1-Tc/Th
.17=1-Tc/Th

However for Th we have to use Kelvin. So we add 273.15 to the given temperature, 425 deg C. Giving us 698.15 deg K. So we sub that in for Th.

.17= 1-Tc/698.15

Then solve for Tc.

(1-.17)*698.15= 579.46 deg K. Subtract 273.15 from 579.46 if you want deg C.

B. Same equation but backwards.

Efficiency=1-Tc/Th

Convert given values to Kelvin and solve

Eff=1-((150+273.15)/(425+273.15))

Eff=0.393 or 39.3%

Answer 2

The cold reservoir temperature of the Carnot engine is approximately 352.75°C, and the maximum efficiency of the steam engine with a cold reservoir temperature of 150°C is approximately 0.83.

Step-by-step explanation:

Overall efficiency:

The maximum efficiency of a heat engine can be calculated using the Carnot efficiency formula:

Carnot Efficiency = 1 - (Tcold / T)

Given the hot steam temperature of 425°C and an efficiency of 17%, we can use these values to find the cold reservoir temperature:

Carnot Efficiency = 1 - (Tcold / 425)

0.17 = 1 - (Tcold / 425)

Tcold = 425 - 0.17 x 425

Tcold = 425 - 72.25

Tcold = 352.75°C

B:

To find the maximum efficiency of the steam engine with a cold reservoir temperature of 150°C, we can use the Carnot efficiency formula:

Carnot Efficiency = 1 - (150 / T)

Efficiency = 1 - (150 / T)

0.17 = 1 - (150 / T)

(150 / T) = 1 - 0.17

(150 / T) = 0.83

T = 150 / 0.83

T = 180.72°C

Therefore, the cold reservoir temperature of the Carnot engine would be approximately 352.75°C, and the maximum efficiency of the steam engine with a cold reservoir temperature of 150°C would be approximately 0.83.