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)

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

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