Chapter 15.3, Problem 28SSC

Interpretation Introduction

Interpretation:

The amount of heat released in freezing 0.25 mol of water is to be calculated.

Concept introduction:

The heat absorbed or released by a substance is dependent on three factors:

1. The mass of the substance.
2. The nature of the substance, its specific heat capacity.
3. The change in temperature of the substance.

The unit of heat is Joules (J), specific heat capacity is J/(kg degree Celsius), and that of mass is kg.

Answer to Problem 28SSC

The mathematical formula to calculate the heat absorbed or released by a substance when the temperature changes is,

$Q=m\times c\times \Delta T$

Where,

$Q$ is the heat released or absorbed,

$m$ is the mass of the substance,

$c$ is the specific heat of the substance,

$\Delta T$ is the difference between the initial and the final temperatures.

Given,

$\begin{array}{l}N=0.25mol\\ N=\frac{mass}{mas{s}_{1mol}}\end{array}$

Molar mass of ${\mathrm{H}}_{2}O=2\times 1+16=18g$

$\begin{array}{c}\mathrm{mass}=N\times mas{s}_{1mol}\\ =0.25\times 18\\ =4.5g\\ \Delta T=0°C-25°C\\ =-25°C\\ \mathrm{Q}=m\times c\times \Delta \mathrm{T}\\ Q=4.5\times 4.19\times -25\\ Q=-471.375\mathrm{J}\end{array}$

Explanation of Solution

The heat released is calculated in the following manner:

1. Calculate the molar mass of water
2. Find the mass of water that contains the given amount of moles
3. Calculate the change in temperature
4. Calculate the heat released using the formula,

$Q=m\times c\times \Delta T$

Conclusion

The amount of heat released is -471.375J.