An engine is supposed to operate between two | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The maximum efficiency of a heat engine is given by the Carnot efficiency formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the abso

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(A) The maximum efficiency of a heat engine is given by the Carnot efficiency formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the absolute temperatures in Kelvin.Given: $T_1 = 727^{\circ}C = 727 + 273 = 1000 \ K$ and $T_2 = 227^{\circ}C = 227 + 273 = 500 \ K$.Substituting the values into the formula:$\eta = 1 - \frac{500}{1000} = 1 - 0.5 = 0.5 = 1/2$.

An engine is supposed to operate between two reservoirs at temperatures $727^{\circ}C$ and $227^{\circ}C$. The maximum possible efficiency of such an engine is:

Similar Questions

$CrO_3$ dissolves in aqueous $NaOH$ to give:

Find the derivative of the function: $(x+\cos x)(x-\tan x)$

What is the product formed when ethyl alcohol reacts with chlorine in the presence of $NaOH$?

The incentre of the triangle with vertices $(1, \sqrt{3})$,$(0, 0)$ and $(2, 0)$ is

The temperature at which a real gas obeys ideal gas laws over an appreciable range of pressure is called:

Vedclass Test Series

Exam Paper Generator

Online Exam Module