Answer to Question #178365 in Molecular Physics | Thermodynamics for Roohi

Question

A resistor initially at 300k has a het capacity of 10Jk^-1 it is thermally insulted from its surrounding and is connected to a battery from which it takes 1200J through an electric current calculate the change in entropy of the resistor and the surrounding.

Answer 1

Solution.

$T_1=300K;$

$C=10JK^{-1};$

$\Delta Q=1200J;$

$\Delta S=\dfrac{\Delta Q}{T};$

$C=\dfrac{\Delta Q}{\Delta T}\implies \Delta T=\dfrac{\Delta Q}{C};$

$\Delta T=\dfrac{1200J}{10JK^{-1}}=120K;$

$T_2=T_1+\Delta T;$

$T_2=300K+120K=420K;$

$\Delta S=\dfrac{1200J}{420K}=2.86JK^{-1};$

Answer:$\Delta S=2.86JK^{-1},$ entropy of universe remains constant.