Question

A Carnot engine works between the source and the sink with efficiency 40% . How much temperature of the sink be lowered keeping the source temperature constant so that it's efficiency increases by 10%?
ans: 30K

Answer 1

The temperature of the sink be lowered keeping the source temperature constant so that its efficiency increases by 10% is equal to 20k

Given that;

A Carnot engine works between the source and the sink with an efficiency of 40%.

T2 = 300K ...(1)

To find;

The temperature of the sink be lowered keeping the source temperature constant so that its efficiency increases by 10%

Solution;

We know that,

The efficiency of a Carnot engine is given by the formula,

η = 1 - $\frac{T2}{T1}$ where T2 = Temperature of sink and T1 = Temperature of source

So, η = 1 -  $\frac{T2}{T1}$ = 1 - $\frac{300}{T1}$

0.4 = 1 -  $\frac{300}{T1}$

T1 = $\frac{300}{1 - 0.4}$ = 500k = Temperature of the source

Now according to the question, the New efficiency after a 10% increase will be,

η=0.44 (44%). Therefore,

0.44 = 1 - $T2^{^{'} }$/T1

$\frac{T2^{'} }{T1}$ = 1 - 0.44 = 0.56

$T2^{'}$ = T1 x 0.56 = 500 x 0.56 = 280k ...(2)

Subtracting (2) from (1) we get,

T2 - $T2^{'}$ = 300K - 280K = 20 K

Hence,

The temperature of the sink be lowered keeping the source temperature constant so that its efficiency increases by 10% is equal to 20k

#SPJ1

Answer 2

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

I OVE YOU BEBE

V