Chapter 11, Problem 13SAQ

Interpretation Introduction

Interpretation:

The final temperature of the mixture of $21.4\text{g}$ of ice and $75.3\text{g}$ of water is to be calculated.

Concept Introduction: The final temperature of the mixture depends upon the combination of both the substances. Final temperature of the water is calculated by the formula,

$m_{\text{ice}}\times c_{\text{ice}}\times \Delta T_{\text{ice}}=-m\times c\times \Delta T_{\text{water}}$

To determine: The option corresponding to the final temperature of the mixture.

Correct answer: The correct option is (b).

Reason for correct answer:

Mass of ice is $21.4\text{g}$ .
Mass of water is $75.3\text{g}$ .
Molar mass of ice is $18\text{g}/\text{mol}$ .
The value of specific heat of water, $c$ is $4.184\text{J}/\text{g}°\text{C}$
The molar heat of fusion for ice, $\Delta H_{\text{fusion}}$ is $6.01\text{kJ}/\text{mol}$ .

The energy required for conversion of $21.4\text{g}$ of ice into water is calculated by the formula,

$q=\frac{\text{mass}\text{of}\text{ice}}{\text{molar}\text{mass}\text{of}\text{ice}}\times \Delta H_{\text{fusion}}$          .........(1)

Substitute the value of mass, molar mass of ice and $\Delta H_{\text{fusion}}$ in equation (1).

$\begin{array}{c}q=\frac{21.4\text{g}}{18\text{g}/\text{mol}}\times 6.01\text{kJ}/\text{mol}\\ =7.14\text{kJ}\end{array}$

Temperature, $T_2$ is calculated by the use of specific heat equation,

$\left(T_2-T_1\right)=\frac{q}{mc}$          ..........(2)

Substitute the value of $T_2,q,m\text{and}c$ in equation (2).

$\begin{array}{c}\left(T_2-0.00°\text{C}\right)=\frac{7140\text{J}}{75.3\text{g}\times \text{4}\text{.184}\text{J}/\text{g}°\text{C}}\\ T_2=\frac{7140\text{J}}{315.055\text{J}}\\ T_2=22.7°\text{C}\end{array}$

$\begin{array}{c}\text{Temperature}\text{of}\text{water}\text{acquired}\text{fom}\text{ice}=\text{temperature}\text{of}\text{water}\text{at}\text{room}\text{temperature}-22.7°\text{C}\\ =\text{55}\text{.3}°\text{C-22}\text{.7}°\text{C}\\ =\text{32}\text{.6}°\text{C}\end{array}$

It is assumed that heat is not releases into surrounding.

Final temperature of the water is calculated by the formula,

$m_{\text{ice}}\times c_{\text{ice}}\times \Delta T_{\text{ice}}=-m\times c\times \Delta T_{\text{water}}$                ........(3)

Substitute the values of $m_{\text{ice}},c_{\text{ice}},\Delta T_{\text{ice}},m,c,\Delta T_{\text{water}}$ in equation (3).

$\begin{array}{c}21.4\text{g}\times 4.184\text{J}/\text{g}°\text{C}\times \left(T_{\text{final}}-0.00\right)=-(75.3\text{g}\times 4.184\text{J}/\text{g}°\text{C}\times (T_{\text{final}}-32.6°\text{C}))\\ 21.4\text{g}\times T_{\text{final}}=-75.3\text{g}\times (T_{\text{final}}-32.6°\text{C})\\ T_{\text{final}}=25.4°\text{C}\end{array}$

Therefore, the final temperature of the mixture is $25.4°\text{C}$ .

Reason for incorrect options:

All the other given values in option (a), (c) and (d) do not match with the above calculated value of final temperture. Hence, they are incorrect.