Question

A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6. When T2 is lowered by 62K its efficiency increases to 1/3.Then T1 and T2 are , respectively
1)330K and 268K 2)310K and 248K 3)372K and 310K 4)372K and 330K

Answer 1

Answer:

Hey mate your answer is here

That is T1= 372

T2=310

Answer 2

Given that,

Efficiency $\eta=\dfrac{1}{6}$

We need to calculate the value of T₂

Using formula of the efficiency of carnot engine

$\eta=(1-\dfrac{T_{2}}{T_{1}})$

Put the value into the formula

$\dfrac{1}{6}=(1-\dfrac{T_{2}}{T_{1}})$

$\dfrac{1}{6}-1=-\dfrac{T_{2}}{T_{1}}$

$\dfrac{T_{2}}{T_{1}}=-\dfrac{1}{6}+1$

$\dfrac{T_{2}}{T_{1}}=\dfrac{5}{6}$

$T_{2}=\dfrac{5}{6}T_{1}$...(I)

When T₂  is lowered by 62K its efficiency increases to 1/3.

We need to calculate the value of T₁

Using formula of efficiency

$\eta=1-\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$\dfrac{1}{3}=1-\dfrac{T_{2}-62}{T_{1}}$

$\dfrac{1}{3}-1=-\dfrac{T_{2}-62}{T_{1}}$

$\dfrac{T_{2}-62}{T_{1}}=\dfrac{2}{3}$

Put the value of T₂

$\dfrac{\dfrac{5}{6}T_{1}-62}{T_{1}}=\dfrac{2}{3}$

$\dfrac{5}{6}T_{1}-\dfrac{2}{3}T_{1}=62$

$\dfrac{5T_{1}-4T_{1}}{6}=62$

$T_{1}=62\times6$

$T_{1}=372\ K$

Now, Put the value of T₁ in equation (I)

$T_{2}=\dfrac{5}{6}\times372$

$T_{2}=310\ K$

Hence, The value of T₁ and T₂ is 372 K and 310 K.

(3) is correct option

Learn more :

Topic : efficiency

https://brainly.in/question/8539732