Answer to Question #131219 in Molecular Physics | Thermodynamics for Ankita

Question

the gain of entropy is 7.24kcal/kg when 5kg of water at 100°c is converted to steam at same temperature. find quantity of heat absorbed

Answer 1

solution

given data

Antropy (s)=7.24Kcal/k

$s=7.24\times4.18KJ/k$

$=$ $30.26KJ/k$

tempreture(T)=100⁰C

=373.15k

antropy for a reversible process can be written as

$\Delta s=\frac{\Delta Q}{T}$

where $\Delta Q$ is heat absorbed during procces

T is tempreture and $\Delta s$ is the change in antropy.

Q can be written as

$\Delta Q=\Delta sT$

by putting the value of s and T

$Q=30.26\times 373.15$

$\fcolorbox{green}{yellow}{$Q=11291.5KJ$}$

therefore quantity of heat absorbed is 11291.5KJ .