Question

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4°C. What is the increase in entropy (in J/K) of the car due to this heat transfer alone?

On a winter day, a certain house loses 5.80 ✕ 108 J of heat to the outside (about 550,000 Btu). What is the total change in entropy (in J/K) due to this heat transfer alone, assuming an average indoor temperature of 23.5°C and an average outdoor temperature of 5.30°C?

Answer 1

a)
`\Delta s=443037.9747\ J.K^(-1)`

b)
`\Delta s=31868131.8681\ J.K^(-1)`

Step-by-step explanation:

a)

Given:

amount of heat transfer occurred,
`dQ=3.5* 10^6\ J`

initial temperature of car,
`T_i=36.5+273=309.5\ K`

final temperature of car,
`T_f=44.4+273=317.4\ K`

We know that the change in entropy is given by:

`\Delta s=(dQ)/(T_f-T_i)`

`\Delta s=(3.5* 10^6)/((44.4-36.5))` (heat is transferred into the system of car)

`\Delta s=443037.9747\ J.K^(-1)`

b)

amount of heat transfer form the system of house,
`dQ=5.8* 10^8\ J`

initial temperature of house,
`T_i=23.5+273=296.5\ K`

final temperature of house,
`T_f=5.3+273=278.3\ K`

`\Delta s=(dQ)/(T_f-T_i)`

`\Delta s=(5.8* 10^8)/(278.3-296.5)`

`\Delta s=31868131.8681\ J.K^(-1)`