Chapter 12, Problem 35PS

Interpretation Introduction

Interpretation:

The energy supply has to be calculated.

Concept introduction:

The energy can be calculated by using following formula,

$\text{q =}\text{m × c × ΔT }$

Answer to Problem 35PS

The total energy supply is  $\text{1}\text{.8×}{\text{10}}^{\text{4}}\text{kJ}$

Explanation of Solution

$\begin{array}{l}\text{The energy can be calculated by using following formula, }\\ \text{q =}\text{m × c × ΔT }\\ \text{Mass (m)}\text{=}\text{12}\text{.0 Kg}\\ \text{c}\text{=}\text{4}\text{.7}\text{J/g}\text{.K}\\ \text{ΔT }\text{=}\text{-33}\text{.3-(-50}\text{.0)}\text{=}\text{16}\text{.7}{}^{\text{o}}\text{C}\text{=}\text{16}\text{.7}\text{K}\\ \text{q =}\text{m × c × ΔT }\\ \text{q =}\text{12g × 4}\text{.7}\text{J/g}\text{.K × 16}\text{.7}\text{K}\text{=}\text{941}\text{.88J }\\ \text{q =}\text{941}\text{.88J }\text{(or)}\text{9}\text{.41}\text{.88×}{\text{10}}^{\text{2}}\text{kJ}\end{array}$

 Liquid at -33.3 ^oC gets converted to vapor at -33.3 ^oC when latent heat is supplied

  $\text{Heat supplied = mass × heat of vaporization}$

 $\begin{array}{l}\text{Mass (m)}\text{=}\text{12}\text{.0 Kg}\\ \text{Number of moles}\text{=}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}\text{=}\frac{\text{12}\text{.0 Kg}}{\text{17}}\\ \text{Number of moles}\text{=}\text{0}\text{.70588}\\ \text{the}\text{enthalpy}\text{of}\text{vaporization}\text{is}\text{23}\text{.33}\text{kJ/mol}\\ \text{heat}\text{supplied}\text{=}\text{number}\text{of}\text{moles}\text{×}\text{enthalpy}\text{of}\text{vaporization}\\ \text{heat}\text{supplied}\text{=}\text{0}\text{.70588×}\text{23}\text{.33}\text{kJ/mol}\\ \text{heat}\text{supplied}\text{=}\text{1}\text{.646×}{\text{10}}^{\text{4}}\text{kJ}\end{array}$

$\begin{array}{l}\text{The energy can be calculated by using following formula, }\\ \text{q =}\text{m × c × ΔT }\\ \text{Mass (m)}\text{=}\text{12}\text{.0 Kg}\\ \text{c}\text{=}\text{2}\text{.2}\text{J/g}\text{.K}\\ \text{ΔT }\text{=}\text{0-(-33}\text{.3)}\text{=}\text{33}\text{.3}{}^{\text{o}}\text{C}\text{=}\text{33}\text{.3}\text{K}\\ \text{q =}\text{m × c × ΔT }\\ \text{q =}\text{12g × 2}\text{.2}\text{J/g}\text{.K × 33}\text{.3}\text{K}\text{=}\text{879}\text{.12J }\\ \text{q =}\text{879}\text{.12J }\text{(or)}\text{8}\text{.79×}{\text{10}}^{\text{2}}\text{kJ}\end{array}$

$\begin{array}{l}\text{The total energy supplied is given below,}\\ \text{Total}\text{energy}\text{supplied}\text{=}\text{9}\text{.41×}{\text{10}}^{\text{2}}\text{kJ + 1}\text{.646×}{\text{10}}^{\text{4}}\text{kJ}\text{+}\text{8}\text{.79×}{\text{10}}^{\text{2}}\text{kJ }\\ \text{Total}\text{energy}\text{supplied}\text{=}\text{1}\text{.8×}{\text{10}}^{\text{4}}\text{kJ}\end{array}$

Conclusion

The total energy supply was calculated.