   Chapter 12, Problem 39PS

Chapter
Section
Textbook Problem

Liquid ammonia, NH3(ℓ), was once used in home refrigerators as the heat transfer fluid. The specific heat capacity of the liquid is 4.7 J/g · K and that of the vapor is 2.2 J/g · K. The enthalpy of vaporization is 23.33 kJ/mol at the boiling point. If you heat 12 kg of liquid ammonia from −50.0 °C to its boiling point of −33.3 °C, allow it to evaporate, and then continue warming to 0.0 °C, how much energy must you supply?

Interpretation Introduction

Interpretation:

The energy supply has to be calculated.

Concept introduction:

The energy can be calculated by using following formula,

q =m × c × ΔT

Explanation

The energy can be calculated by using following formula, q =m × c × ΔT Mass (m)=12.0 Kgc=4.7J/g.KΔT =-33.3-(-50.0)=16.7oC=16.7Kq =m × c × ΔT q =12g × 4.7J/g.K × 16.7K=941.88J q =941.88J (or)9.41.88×102 kJ

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

Heat supplied = mass × heat of vaporization

Mass (m)=12.0 KgNumber of moles=MassMolar massNumber of moles=12.0 Kg17Number of moles=0.70588theenthalpyofvaporizationis23.33kJ/molheatsupplied=numberofmoles×enthalpyofvaporizationheatsupplied=0

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