Question

A hot rock ejected from a volcano’s lava fountain cools from 1100º C to 40.0º C, and its entropy decreases by 950 J/K. How much heat transfer occurs from the rock?

Short answer

The rock transfers 800.85 kJ of heat.

Step-by-step solution

Introduction

Entropy can be defined as the amount of heat that has converted to unuseful forms per unit temperature, at a given temperature.

Here,

\[\Delta S= \frac{Q}{{\bf{T}}}\]

\[\Delta S\] - change in entropy.

Q- heat transferred.

T - temperature.

 Given parameters and formula for change in entropy

Entropy change = 950 J/K

Temperature final T_c=40 ̊ C =313 K

Temperature initial T_h =1100 ̊ C

Change in entropy

Here,

\[\Delta S\] - change in entropy.

Q - heat transferred.

T - temperature.

 Calculate the average temperature

Average of the two temperatures

\begin{aligned}{l}T= \frac{{({T_h} + {T_c})}}{2}\\T= 843 K\end{aligned}

 Calculate heat transfer

Heat transfer occurring in rock is

\begin{aligned}Q= T \times \Delta S\\= 843\;{\rm{K}} \times 950 {{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{K}}}}\right\{\rm{K}}}\\= 800.85 {\rm{kJ}}\end{aligned}

Therefore, the heat transfer of 800.85kJ occurs from the rock.