Chapter 9, Problem 35QRT

Interpretation Introduction

Interpretation:

The total heat energy required to change $0.5$ mole of ice at $-5^{\circ }C$ to $0.5$ mole of steam at ${100}^{\circ }C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $\Delta H^0=m\times s\times \Delta \theta$

When two phases are different, $\Delta H^0=n\times {\Delta }_{fusorvap}{H}^0$

Answer to Problem 35QRT

The total heat energy required to change $0.5$ mole of ice at $-5^{\circ }C$ to $0.5$ mole of steam at ${100}^{\circ }C$  is $27kJ$.

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06J{g}^{-1}{}^{\circ }{C}^{-1}$ and $4.184J{g}^{-1}{}^{\circ }{C}^{-1}$ respectively. The fusion of enthalpy for solid ice is $6.02kJ/mol$. The vaporization enthalpy of liquid water is $40.7kJ/mol$.

Calculation of mass of ice:

  No. of moles $\left(n\right)=\frac{m}{M}\Rightarrow m=n\times M=0.5mol\times 18.0152\frac{g}{mol}=9.0g.$

1. 1. Heat the ice from $-5^{o}C$ to $0^{o}C$

The heat energy required for $9.0g$ of ice to melt can be calculated as given below.

  $\begin{array}{c}\Delta H^0=m\times s\times \Delta \theta \\ \\ =\left(9g\right)\left(2.06J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left[\left(0^{\circ }C\right)-\left(-5^{\circ }C\right)\right]\\ \\ =92.7J=92.7J\times \frac{1kJ}{1000J}=0.0927kJ.\end{array}$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $\Delta \theta =$ Temperature difference.

1. 2. Melt the ice at $0^{o}C$

The heat energy required for this process can be calculated as given below.

  $\begin{array}{c}\Delta H^0=n\times {\Delta }_{fus}{H}^0\\ \\ =0.5mol\times 6.020\frac{kJ}{mol}=3.01kJ.\end{array}$

Where, $n=$ No. of moles of ice and ${\Delta }_{fus}{H}^0=$ Fusion enthalpy of solid ice.

1. 3. Heat the water from $0^{o}C$ to $100^{o}C$

The heat energy required for this process can be calculated as given below.

  $\begin{array}{c}\Delta H^0=m\times s\times \Delta \theta \\ \\ =\left(9g\right)\left(4.184J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left[\left({100}^{\circ }C\right)-\left(0^{\circ }C\right)\right]\\ \\ =3765.6J=3765.6J\times \frac{1kJ}{1000J}=3.7656kJ/mol.\end{array}$

Where, $m=$ mass of water, $s=$ Specific heat of liquid water and $\Delta \theta =$ Temperature difference.

1. 4. Boil the water at $100^{o}C$

The heat energy required for this process can be calculated as given below.

  $\begin{array}{c}\Delta H^0=n\times {\Delta }_{vap}{H}^0\\ \\ =0.5mol\times 40.7\frac{kJ}{mol}=20.35kJ.\end{array}$

Where, $n=$ No. of moles of water and ${\Delta }_{vap}{H}^0=$ Enthalpy of vaporization of liquid water.

Now, the total heat energy required to change $0.5$ mole of ice at $-5^{\circ }C$ to $0.5$ mole of steam at ${100}^{\circ }C$ is the sum total of all the heat energies required for each transition:

  $\begin{array}{c}\Delta H^{\circ }=0.0927kJ+3.01kJ+3.7656kJ+20.35kJ\\ \\ =27.2183kJ\cong 27kJ.\end{array}$

Therefore, the total heat energy required to change $0.5$ mole of ice at $-5^{\circ }C$ to $0.5$ mole of steam at ${100}^{\circ }C$ has been calculated to be $27kJ$.