Chapter 10, Problem 24A

Interpretation Introduction

Interpretation: Required heat to raise the temperature of the same sample of metal from $\text{25}°\text{C}$ to $40°\text{C}$ needs to be determined.

Concept Introduction: The expression for heat is as follows:

  $Q=s\times m\times \Delta T$

Here, Q represent energy required, s denotes specific heat capacity, m represent mass of the sample in grams and $\Delta T$ represent change in temperature.

Answer to Problem 24A

Required heat is $25.2\text{kJ}$ .

Explanation of Solution

The sample has the same mass in both cases.

Calculate temperature change for 1^st case as follows:

  $\begin{array}{c}\Delta T_1=20°\text{C}-15°\text{C}\\ =5°\text{C}\end{array}$

Calculate temperature change for 2nd case as follows:

  $\begin{array}{c}\Delta T_1=40°\text{C}-25°\text{C}\\ =15°\text{C}\end{array}$

Hence,

  $Q_1=s\times m\times \Delta T_1$ …… (1)

and $Q_2=s\times m\times \Delta T_2$ …… (2)

Divide equation (1) by (2).

  $\frac{Q_1}{Q_2}=\frac{s\times m\times \Delta T_1}{s\times m\times \Delta T_2}$

  $\frac{Q_1}{Q_2}=\frac{\Delta T_1}{\Delta T_2}$ …… (3)

Given values are as follows:

  $\begin{array}{l}{Q}_1=8.40\text{kJ}\\ \Delta T_1=5.0°\text{C}\\ \Delta T_2=15.0°\text{C}\\ Q_2=?\end{array}$

Substitute all values in equation (3).

  $\begin{array}{c}\frac{Q_1}{Q_2}=\frac{\Delta T_1}{\Delta T_2}\\ Q_2=\frac{\Delta T_2\times Q_1}{\Delta T_1}\\ =\frac{15°\text{C}\times \text{8}\text{.40}\text{kJ}}{5°\text{C}}\\ =\frac{126}{5}\text{kJ}\end{array}$

  $Q_2=25.2\text{kJ}$

Therefore, required heat is $25.2\text{kJ}$ .