Chapter 15, Problem 98E

The following information is given for bismuth at $1\text{atm}$:

$\text{boiling}\text{point}=1627°\text{C}$ ${\text{ΔH}}_{\text{vap}}\left(1627°\text{C}\right)=822.9\text{J}/\text{g}$

$\text{melting}\text{point}=271.0°\text{C}$ ${\text{ΔH}}_{\text{fus}}\left(271.0°\text{C}\right)=52.60\text{J}/\text{g}$

$\text{specific}\text{heat}\text{gas}=0.1260\text{J}/\text{g}\cdot °\text{C}$

$\text{specific}\text{heat}\text{liquid}=0.1510\text{J}/\text{g}\cdot °\text{C}$

A $22.80\text{-g}$ sample of liquid bismuth at $553.0°\text{C}$ is poured into a mold and allowed to cool to $28.0°\text{C}$. What quantity of energy (in kilojoules) is transferred to the atmosphere in this process?

Interpretation Introduction

Interpretation:

The amount of energy released by the temperature of liquid bismuth is changed from $553°\text{C}$ to $28.0°\text{C}$ is to be calculated.

Concept introduction:

The amount of energy required to change the state of a substance is known as enthalpy. It is the different in the energy of final and initial state of a substance. The negative and positive sign of enthalpy indicates the energy released and energy absorbed, respectively, during the phase change.

Answer to Problem 98E

The amount of energy released by the temperature of liquid bismuth is changed from $553°\text{C}$ to $28.0°\text{C}$ is $-2.85\text{kJ}$.

Explanation of Solution

Bismuth solidifies at $271.0°\text{C}$. Hence, the temperature of bismuth is first changed from $553°\text{C}$ to $271.0°\text{C}$.

The amount of energy released when temperature changes from $553°\text{C}$ to $271.0°\text{C}$ is calculated by the formula shown below.

${\text{q}}_{\text{1}}=\text{mc}\left({\text{T}}_2-{\text{T}}_1\right)$…(1)

Where,

• $\text{m}$ is the mass of the sample.

• $\text{c}$ is the specific heat of bismuth.

• ${\text{T}}_{\text{2}}$ is the final temperature $\left(271.0°\text{C}\right)$.

• ${\text{T}}_{\text{1}}$ is the initial temperature $\left(553°\text{C}\right)$.

The specific heat of bismuth is $0.1510\text{J}/\text{g}\cdot °\text{C}$.

Substitute the mass, final, initial temperature and specific heat of zinc in equation (1).

$\begin{array}{rll}{\text{q}}_{\text{1}}& =& \text{22}\text{.80}\text{g}\times 0.1510\text{J}/\text{g}\cdot °\text{C}\left(271.0°\text{C}-\left(553°\text{C}\right)\right)\\ & =& -970.87\text{J}\end{array}$

The amount of energy required for phase transformation is calculated by the formula shown below.

${\text{q}}_{\text{2}}=\text{m}\Delta {\text{H}}_{\text{fus}}$…(2)

Where,

• is the heat of fusion.

The heat of fusion of bismuth is $-52.60\text{J}/\text{g}$.

Substitute the mass and heat of fusion in equation (2).

$\begin{array}{rll}{\text{q}}_{\text{2}}& =& \text{22}\text{.80}\text{g}\times -52.60\text{J}/\text{g}\\ & =& -1199.28\text{J}\end{array}$

The amount of energy required to change the temperature of bismuth from $271.0°\text{C}$ to $28.0°\text{C}$ is calculated by the formula shown below.

${\text{q}}_{\text{3}}=\text{mc}\left({\text{T}}_2-{\text{T}}_1\right)$…(3)

Where,

• $\text{m}$ is the mass of the sample.

• $\text{c}$ is the specific heat of bismuth.

• ${\text{T}}_{\text{2}}$ is the final temperature $\left(28.0°\text{C}\right)$.

• ${\text{T}}_{\text{1}}$ is the initial temperature $\left(271.0°\text{C}\right)$.

The specific heat of bismuth is $0.123\text{J}/\text{g}\cdot °\text{C}$.

Substitute the mass, final, initial temperature and specific heat of bismuth in equation (3).

$\begin{array}{rll}{\text{q}}_{\text{3}}& =& \text{22}\text{.80}\text{g}\times 0.123\text{J}/\text{g}\cdot °\text{C}\left(28.0°\text{C}-\left(271.0°\text{C}\right)\right)\\ & =& -681.47\text{J}\end{array}$

The amount of energy released by the temperature of liquid bismuth is changed from $553°\text{C}$ to $28.0°\text{C}$ is calculated as shown below.

$\begin{array}{rll}\text{q}& =& {\text{q}}_1+{\text{q}}_2+{\text{q}}_3\\ & =& -970.87\text{J}+\left(-1199.28\text{J}\right)+\left(-681.47\text{J}\right)\\ & =& -2851.62\text{J}\end{array}$

Convert $-2851.62\text{J}$ to $\text{kJ}$.

$\begin{array}{rll}-2851.62\text{J}& =& \frac{-2851.62}{1000}\text{kJ}\\ & =& -2.85\text{kJ}\end{array}$

Conclusion

The amount of energy released by the temperature of liquid bismuth is changed from $553°\text{C}$ to $28.0°\text{C}$ is $-2.85\text{kJ}$.