Question

On a hot summer day, 4.00 ×10^{6} J of heat transfer into a parked car takes place, increasing its temperature from 35.0ºC to 45.0ºC . What is the increase in entropy of the car due to this heat transfer alone? (answer in ×10^{4} J/K)

Answer 1

the change in entropy is ΔS= 1.278*10^4 J/K

Step-by-step explanation:

according to the second law of thermodynamics, and assuming a reversible process

ΔS= ∫dQ/T

where

ΔS= change in entropy

Q= heat exchanged

T = absolute temperature

for sensible heat

Q= m*c*(T-To)

c= specific heat , T= final temperature , To= initial temperature

therefore

dQ= m*c*dT

ΔS= ∫dQ/T = m*c*∫dT/T = m*c*ln (T2/T1) = Q* ln (T2/T1)/(T2-T1)

replacing values

ΔS= Q* ln (T2/T1)/(T2-T1) = 4.00 ×10^6 J* ln(318 K/308 K)/(318 K-308 K) =12780.64 J/K

ΔS= 1.278*10^4 J/K