Question

A Carnot engine whose low-temperature reservoir is at 19.5°C has an efficiency of 23.0%. By how much should the Celsius temperature of the high-temperature reservoir be increased to increase the efficiency to 64.1%?

Answer 1

Answer:

434.3727 °C

Explanation:

Given :

The low temperature reservoir, TL = 19.5 °C

The conversion of T( °C) to T(K) is shown below:  

T(K) = T( °C) + 273.15    

So,  

TL = (19.5 + 273.15) K = 292.65 K  

Given: E₁ = 23.0 %

Let the temperature of the gas is , TH₁ = x K  

The engine's efficiency of a Carnot engine is:

$Carnot's\ Efficiency=\frac {T_H-T_L}{T_H}\times 100 \%$

So,

$23.0 \%=\frac {(x)-292.15}{x}\times 100 \%$

x = 379.4156 K

Now,

Given: E₂ = 64.1 %

Let the temperature of the gas is , TH₂ = y K  

The engine's efficiency of a Carnot engine is:

$Carnot's\ Efficiency=\frac {T_H-T_L}{T_H}\times 100 \%$

So,

$64.1 \%=\frac {(y)-292.15}{y}\times 100 \%$

y = 813.7883 K

Also,

The conversion of T(K)  to T( °C)is shown below:  

T( °C) = T(K)  - 273.15  

x = 379.4156 K = 106.2656 °C

y = 813.7883 K = 540.6383 °C

The temperature that must be increased = 540.6383 °C - 106.2656 °C = 434.3727 °C