Chapter 9, Problem 98QRT

Interpretation Introduction

Interpretation:

The total heat energy transfer required to raise the temperature of $10kg$ liquid ammonia from $-{50}^{\circ }C$ to $0^{\circ }C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid, $\Delta H^0=m\times s\times \Delta \theta$

Where, $m=$ mass of substance, $s=$ Specific heat of substance in a particular phase and $\Delta \theta =$ Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid, $\Delta H^0=n\times {\Delta }_{fusorvap}{H}^0$

Where, $n=$ No. of moles of substance , ${\Delta }_{fus}{H}^0=$ Fusion enthalpy of a substance in its solid phase and ${\Delta }_{vap}{H}^0=$ enthalpy of vaporization of a substance in its liquid phase.

Answer to Problem 98QRT

The total heat energy transfer required to raise the temperature of $10kg$ liquid ammonia from $-{50}^{\circ }C$ to $0^{\circ }C$ is $1.6\times 10^{4}kJ.$

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are $2.2J{g}^{-1}{}^{\circ }{C}^{-1}$ and $4.7J{g}^{-1}{}^{\circ }{C}^{-1}.$ respectively. Its normal boiling point is $-{33.4}^{\circ }C.$  The vaporization enthalpy is $23.5kJ/mol.$

Calculation of no. of moles of ice:

No. of moles $\left(n\right)=\frac{m}{M}=\frac{10kg}{17.0304g/mol}\times \frac{1000g}{1kg}=5.9\times {10}^{2}mol.$

1. 1. Heat it from $-50^{o}C$ to its boiling point $-33.4^{o}C$

The heat energy absorbed during this process can be calculated as given below.

  $\begin{array}{c}\Delta H^0=m\times s\times \Delta \theta \\ \\ =\left(10kg\right)\left(4.7J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left(\frac{1kJ}{1000J}\right)\left[\left(-{33.4}^{\circ }C\right)-\left(-{50}^{\circ }C\right)\right]\\ \\ =780kJ.\end{array}$

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

1. 2. Vaporize the liquid to gas at $-33.4^{o}C$

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

  $\begin{array}{c}\Delta H^0=n\times {\Delta }_{vap}{H}^0\\ \\ =5.9\times {10}^{2}mol\times 23.5\frac{kJ}{mol}=14000kJ.\end{array}$

1. 3. Heat it further from $-33.4^{o}C$ to $0^{o}C$

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

  $\begin{array}{c}\Delta H^0=m\times s\times \Delta \theta \\ \\ =\left(10kg\right)\times \left(\frac{1000g}{1kg}\right)\left(0.97J{g}^{-1}{}^{\circ }{C}^{-1}\right)\times \left(\frac{1kJ}{1000J}\right)\left[\left(0^{\circ }C\right)-\left(-{33.4}^{\circ }C\right)\right]\\ \\ =730kJ.\end{array}$

Now, the total heat energy transfer required to raise the temperature of $10kg$ liquid ammonia from $-{50}^{\circ }C$ to $0^{\circ }C$ is the sum total of all the heat energies absorbed for each transition:

  $\begin{array}{c}\Delta H^{\circ }=780kJ+14000kJ+730kJ\\ \\ =1.6\times {10}^{4}kJ.\end{array}$

Therefore, the total heat energy transfer required to raise the temperature of $10kg$ liquid ammonia from $-{50}^{\circ }C$ to $0^{\circ }C$ is $1.6\times 10^{4}kJ.$