Question

A resistor initially at 300 K has a heat capacity of 10 J K⁻¹ . It is thermally insulated from its surrounding and is connected to a battery from which it takes 1200 J through an electric current. Calculate the change in entropy of (i) the resistor, 

(ii) the surroundings.

Answer 1

Answer :-

Given:-

$\boxed{T_1=300K}$

$\boxed{C=10JK^{-1}}$

$\boxed{ \Delta Q=1200J}$

We know that

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

We get

$\bigstar$

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

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

$T_2=T_1+\Delta T$

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

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

Answer:$\boxed{ \Delta S=2.86JK^{-1}}$ 

. entropy of universe remains constant

$\boxed{\Delta S _{resistor} =- \Delta S{surroundings}}$