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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Liquid ammonia, NH3(), 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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