Chapter 15, Problem 79E

Interpretation Introduction

Interpretation:

The energy released when $2.30\text{kg}$ of gold is cooled from $88\text{°C}$ to $22\text{°C}$ is to be stated.

Concept introduction:

Specific heat is the property of a pure substance. Specific heat is the heat flow required to change the temperature of $1\text{g}$ of a substance by $1$ degree Celsius. It measures the relative ease with which a substance may be heated or cooled. A substance which has a low specific heat gains less amount of energy in warming through a temperature change. A substance which has a high specific heat gains higher amount of energy in warming through a temperature change.

Answer to Problem 79E

The energy released when $2.30\text{kg}$ of gold is cooled from $88\text{°C}$ to $22\text{°C}$ is $-2\times {10}^1\text{kJ}$.

Explanation of Solution

The specific heat, $\text{c}$ is calculated by the formula given below.

$\text{c}=\frac{\text{q}}{\text{m}\times \Delta \text{T}}$…(1)

Where,

• $\text{q}$ is the energy

• $\text{m}$ is the mass.

• is the temperature change.

The value of $\Delta \text{T}$ is calculated by the formula given below.

$\text{ΔT}={\text{T}}_{\text{f}}-{\text{T}}_{\text{i}}$…(2)

Where,

• ${\text{T}}_{\text{f}}$ is the final temperature of the substance.

• ${\text{T}}_{\text{i}}$ is the initial temperature of the substance,

The specific heat of gold $\left(\text{c}\right)$ is $0.13\text{J}/\text{g}\cdot \text{°C}$.

The mass of the gold $\left(\text{m}\right)$ is $2.30\text{kg}$.

The relation between $\text{g}$ and $\text{kg}$ is given below.

$\text{1}\text{kg}=\text{1000}\text{g}$

The probable conversion factors are given below.

$\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{and}\frac{\text{1000}\text{g}}{\text{1}\text{kg}}$

The conversion factor to deermine $\text{g}$ from $\text{kg}$ is given below.

$\frac{\text{1000}\text{g}}{\text{1}\text{kg}}$

Therefore, the mass is calculated below.

$\begin{array}{rll}\text{m}& =& 2.30\text{kg}\times \frac{\text{1000}\text{g}}{\text{1}\text{kg}}\\ & =& 2300\text{g}\end{array}$

The value of ${\text{T}}_{\text{f}}$ is $22\text{°C}$.

The value of ${\text{T}}_{\text{i}}$ is $88\text{°C}$.

Substitute the value of ${\text{T}}_{\text{f}}$ and ${\text{T}}_{\text{i}}$ in equation (2).

$\begin{array}{rll}\text{ΔT}& =& 22\text{°C}-88\text{°C}\\ & =& -66\text{°C}\end{array}$

Substitute the value of $\Delta \text{T}$, $\text{c}$ and $\text{m}$ in equation (1).

$\begin{array}{rll}0.13\text{J}/\text{g}\cdot \text{°C}& =& \frac{\text{q}}{2300\text{g}\times \left(-66\text{°C}\right)}\\ \text{q}& =& 0.13\text{J}/\text{g}\cdot \text{°C}\times 2300\text{g}\times \left(-66\text{°C}\right)\\ & =& -2\times {10}^4\text{J}\end{array}$

The relation between $\text{J}$ and $\text{kJ}$ is given below.

$1\text{kJ}=\text{1000}\text{J}$

The probable conversion factors are given below.

$\frac{1\text{kJ}}{\text{1000}\text{J}}\text{and}\frac{\text{1000}\text{J}}{1\text{kJ}}$

The conversion factor to determine $\text{kJ}$ from $\text{J}$ is given below.

$\frac{1\text{kJ}}{\text{1000}\text{J}}$

Therefore, the amount of energy released is calculated below.

$\begin{array}{rll}\text{q}& =& -2\times {10}^4\text{J}\times \frac{1\text{kJ}}{\text{1000}\text{J}}\\ & =& -2\times {10}^1\text{kJ}\end{array}$

Therefore, the energy released when $2.30\text{kg}$ of gold is cooled from $88\text{°C}$ to $22\text{°C}$ is $-2\times {10}^1\text{kJ}$.

Conclusion

The energy released when $2.30\text{kg}$ of gold is cooled from $88\text{°C}$ to $22\text{°C}$ is $-2\times {10}^1\text{kJ}$.