If a Carnot engine functions at a source temp | vedclass

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

Question

(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.First,convert the temperatures from Celsius to Kelvin:$T_1 = 127 + 273 = 400 	ext{ K}$$T_2 = 87 + 273 = 360 	ext{ K}$Now,substitute these values into the efficiency formula:$\eta = \frac{T_1 - T_2}{T_1} = \frac{400 - 360}{400}$$\eta = \frac{40}{400} = 0.1$To express this as a percentage,multiply by $100$:$\eta = 0.1 	imes 100 = 10\%$

If a Carnot engine functions at a source temperature of $127^{\circ}C$ and a sink temperature of $87^{\circ}C$,what is its efficiency in $\%$?

Similar Questions

What is the common feature of a heat engine based on a cyclic process?

$A$ Carnot engine operates between $300 \ K$ and $600 \ K$. If the work done per cycle is $800 \ J$,then the heat supplied per cycle is ....... $J/cycle$.

An ideal Carnot heat engine has an efficiency of $30\%$. It absorbs heat from a hot reservoir at $727^{\circ}C$. The temperature of the cold reservoir is .... $^{\circ}C$.

An ideal gas heat engine operates in a Carnot cycle between $227^{\circ}C$ and $127^{\circ}C$. It absorbs $6 \times 10^4 \text{ cal}$ of heat at the higher temperature. The amount of heat converted to work is ......... $\times 10^4 \text{ cal}$.

An ideal heat engine exhausting heat at $77^{\circ}C$ is to have a $30\%$ efficiency. It must take heat at ...... $^{\circ}C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module