Chapter 12, Problem 39PS

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

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

TheÂ energyÂ canÂ beÂ calculatedÂ byÂ usingÂ followingÂ formula,Â qÂ =â€‰mÂ Ã—Â cÂ Ã—Â Î”TÂ â€‰MassÂ (m)â€‰=â€‰12.0Â Kgcâ€‰=â€‰4.7â€‰J/g.KÎ”TÂ â€‰=â€‰-33.3-(-50.0)â€‰=â€‰16.7â€‰oCâ€‰=â€‰16.7â€‰KqÂ =â€‰mÂ Ã—Â cÂ Ã—Â Î”TÂ qÂ =â€‰12gÂ Ã—Â 4.7â€‰J/g.KÂ Ã—Â 16.7â€‰Kâ€‰=â€‰941.88JÂ qÂ =â€‰â€‰â€‰941.88JÂ â€‰(or)â€‰9.41.88Ã—â€‰102Â â€‰kJ

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

Â Â HeatÂ suppliedÂ =Â massÂ Ã—Â heatÂ ofÂ vaporization

Â â€‰MassÂ (m)â€‰=â€‰12.0Â KgNumberÂ ofÂ molesâ€‰=â€‰MassMolarÂ massNumberÂ ofÂ molesâ€‰=â€‰â€‰12.0Â Kg17NumberÂ ofÂ molesâ€‰=â€‰0.70588theâ€‰enthalpyâ€‰ofâ€‰vaporizationâ€‰isâ€‰23.33â€‰kJ/molheatâ€‰suppliedâ€‰=â€‰numberâ€‰ofâ€‰molesâ€‰Ã—â€‰enthalpyâ€‰ofâ€‰vaporizationheatâ€‰suppliedâ€‰=â€‰0

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