Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00g$ of ice from $-{10}^{\circ }C$ to ${20}^{\circ }C$ 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 36QRT

The total heat energy required to melt a $36.00g$ of ice from $-{10}^{\circ }C$ to ${20}^{\circ }C$ has been calculated to be $15.7kJ$.

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 no. of moles of ice:

  No. of moles $\left(n\right)=\frac{m}{M}=\frac{36g}{18.0152g/mol}=1.998mol.$

1. 1. Heat the ice from $-10^{o}C$ to $0^{o}C$

The heat energy required for $36g$ of ice to melt can be calculated as given below.

  $\begin{array}{c}\Delta H^0=m\times s\times \Delta \theta \\ \\ =\left(36g\right)\left(2.06J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left[\left(0^{\circ }C\right)-\left(-{10}^{\circ }C\right)\right]\\ \\ =741.6J=741.6J\times \frac{1kJ}{1000J}=0.7416kJ.\end{array}$

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.

  $\begin{array}{c}\Delta H^0=n\times {\Delta }_{fus}{H}^0\\ \\ =1.998mol\times 6.020\frac{kJ}{mol}=12.02796kJ.\end{array}$

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 $20^{o}C$

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

  $\begin{array}{c}\Delta H^0=m\times s\times \Delta \theta \\ \\ =\left(36g\right)\left(4.184J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left[\left({20}^{\circ }C\right)-\left(0^{\circ }C\right)\right]\\ \\ =3012.48J=3012.48J\times \frac{1kJ}{1000J}=3.01248kJ.\end{array}$

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

Now, the total heat energy required to melt a $36.00g$ of ice from $-{10}^{\circ }C$ to ${20}^{\circ }C$ is the sum total of all the heat energies required for each transition:

  $\begin{array}{c}\Delta H^{\circ }=0.7416kJ+12.02796kJ+3.01248kJ\\ \\ =15.78204kJ\cong 15.7kJ.\end{array}$

Therefore, the total heat energy required to melt a $36.00g$ of ice from $-{10}^{\circ }C$ to ${20}^{\circ }C$ has been calculated to be $15.7kJ$.