Question

When 1.50 ✕ 10^5 J of heat transfer occurs into a meat pie initially at 20.0°C, its entropy increases by 465 J/K. What is its final temperature (in degrees)?

Answer 1

The final temperature is 79.16°C.

Step-by-step explanation:

Given that,

Heat
`Q=1.50*10^(5)\ J`

Temperature = 20.0°C

Entropy = 465 J/k

We need to calculate the average temperature

Using relation between entropy and heat

`\Delta S=(\Delta Q)/(T)`

`T=(\Delta Q)/(\Delta S)`

Where, T = average temperature

`\Delta Q`= transfer heat

`\Delta S`= entropy

Put the value into the formula

`T=(1.50*10^(5))/(465)`

`T=322.58\ K`

We need to calculate the final temperature

Using formula of average temperature

`T = (T_(i)+T_(f))/(2)`

`T_(f)=2T-T_(i)`....(I)

Put the value in the equation (I)

`T_(f)=2*322.58-293`

`T_(f)=352.16\ K`

We convert the temperature K to degrees

`T_(f)=352.16-273`

`T_(f)=79.16^(\circ)\ C`

Hence, The final temperature is 79.16°C.