Chapter 3, Problem 3B.5E

Interpretation Introduction

Interpretation:

The entropy change of $\text{1kg}$ aluminum on cooling and also the energy removed as heat on cooling have to be calculated.

Concept Information:

Change in entropy on heating at constant volume: the change in entropy when the temperature is changed from ${\text{T}}_i$ to ${\text{T}}_{\text{f}}$. The greater the heat capacity of substance, the change in entropy increases by increasing temperature. A high heat capacity implies that a lot of heat is required to produce a given change in temperature must be more powerful than for when the heat capacity is low, and the entropy increase is correspondingly high.

$\text{ΔS}\text{=}\text{C}\text{ln}\left(\frac{{\text{T}}_{\text{f}}}{{\text{T}}_{\text{i}}}\right)$

Where,

C is the heat capacity of a substance,

${\text{T}}_{\text{f}}$ is the final temperature of the system,

${\text{T}}_i$ is the initial temperature of the system.