Question

When 1.60 × 10 5 J 1.60×105 J of heat transfer occurs into a meat pie initially at 17.5 °C , 17.5 °C, its entropy increases by 485 J / K . 485 J/K. Estimate the final temperature of the pie.

Answer 1

Step-by-step explanation:

Given

Heat transfer
`Q=1.6* 10^5\ J`

initial Temperature
`T_i=17.5^(\circ)\approx 290.5\ K`

Entropy change
`dS=485\ J/K`

The expression for entropy is given by

`dQ=TdS`

`T=(dQ)/(dS)`

`T=(1.6* 10^5)/(485)`

`T=329.89\ K`

Temperature can be written as average of initial and final temperature

`T=(T_i+T_f)/(2)`

`329.89=(T_f+290.5)/(2)`

`T_f=659.78-290.5`

`T_f=369.28\ K`