Question

An ideal gas heat engine operates in a Carnot engine cycle between 227°C and 127°C it's absorb 6*10power 4 J at high temperature the amount of hear converted into work us

Answer 1

$\huge\mathfrak {Answer:-}$

We know that,

$\frac{W}{Q} = 1 - \frac{T2}{T1}$

Here,

$T1 = (273 + 227)K = 500K$

And,

$T2 = (273 + 127)K = 400K$

So,

$1 - \frac{400}{500} = \frac{1}{5}$

Now,

Efficiency of Heat engine = $\frac{Work \: output}{Heat \: input}$

So,

$\frac{1}{5} = \frac{Work \: output}{6 \times 10^{4} }$

$= \frac{6}{5}{ \times 10^{4} }$

$= 1.2 \times 10 ^{4}cal$

Therefore,

Work output = $1.2 \times 10 ^{4}cal$

$\huge{Be\:Brainly}$❤️

Answer 2

1.2 ten raise to -4........