Chapter 15, Problem 110A

Calculate Suppose that two pieces of iron, one with amass exactly twice the mass of the other, are placed in aninsulated calorimeter. If the original temperatures of thelarger piece and the smaller piece are 90.0°C and 50.0°C,respectively, what is the temperature of the two pieceswhen thermal equilibrium has been established? Referto Table R-9 on page 975 for the specific heat of iron.

Interpretation Introduction

Interpretation: The temperature of the two iron pieces needs to be determined.

Concept introduction: Thermal equilibrium: It is the state where two substances in contact do not exchange any heat energy with each other.

Answer to Problem 110A

The temperature of both iron pieces will be $76.7{}^{\text{o}}\text{C}$ .

Explanation of Solution

Given information:

For,

Iron 1( ${\text{I}}_1$ ) Iron 2( ${\text{I}}_2$ )

  $\begin{array}{l}{\text{m}}_{\text{1}}=\text{m}\\ {\text{m}}_{\text{1}}=\text{2m}\\ {\text{T}}_{\text{1}}={\text{90}}^{\text{o}}\text{C}\\ \text{Specific}\text{heat}=\text{0}\text{.449}\end{array}$

  $\begin{array}{l}{\text{m}}_2=\text{m}\\ {\text{m}}_2=\text{m}\\ {\text{T}}_2=5{\text{0}}^{\text{o}}\text{C}\\ \text{Specific}\text{heat}=\text{0}\text{.449}\end{array}$

Calculation:

The chemical potential energy is absorbed or released during the heating of the chemical processes or reactions.

  $q=c\times m\times \Delta T$

In order to reach equilibrium, the heat lost by one iron piece will be equal to heat gain by other iron pieces. The final temperature will be the same for both iron pieces.

  $\begin{array}{c}-q_1=q_2\\ \text{-m}\times \text{0}\text{.449}\times {\text{(T}}_f-\text{50)}=2\text{m}\times \text{0}\text{.449}\times {\text{(T}}_f-\text{90)}\\ -{\text{(T}}_f-\text{50)}=2{\text{(T}}_f-\text{90)}\\ -{\text{T}}_f+50=2{\text{T}}_f-180\\ T_f\text{=}\frac{\text{230}}{3}{}^{\text{o}}\text{C}\\ =76.7{}^{\text{o}}\text{C}\end{array}$

Conclusion

Equilibrium of any substance is the state where both the equations are in the same state.