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

Answer:

The final temperature is 79.16°C.

Explanation:

Given that,

Heat $Q=1.50\times10^{5}\ J$

Temperature = 20.0°C

Entropy = 465 J/k

We need to calculate the average temperature

Using relation between entropy and heat

$\Delta S=\dfrac{\Delta Q}{T}$

$T=\dfrac{\Delta Q}{\Delta S}$

Where, T = average temperature

$\Delta Q$= transfer heat

$\Delta S$= entropy

Put the value into the formula

$T=\dfrac{1.50\times10^{5}}{465}$

$T=322.58\ K$

We need to calculate the final temperature

Using formula of average temperature

$T = \dfrac{T_{i}+T_{f}}{2}$

$T_{f}=2T-T_{i}$....(I)

Put the value in the equation (I)

$T_{f}=2\times322.58-293$

$T_{f}=352.16\ K$

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.