Liquid ammonia can be used as a refrigerant and heat transfer fluid. Given the f

Question

Liquid ammonia can be used as a refrigerant and heat transfer fluid. Given the following information

boiling point

-33.4°C

specific heat capacity, NH  3  (  l  )

4.7 J/(g·  o  C)

heat of vaporization

1380 J/g

specific heat capacity, NH  3  (  g  )

2.2 J/(g·  o  C)

i) In what phase will ammonia be at -60.0oC, solid(S), liquid (L), or gas(G)? ___
ii) In what phase will ammonia be at -10.0oC, solid(S), liquid (L), or gas(G)? ___

{ NOTE: Give all answers for the following in standard form. Do NOT use scientific notation. Also remember to use the rules of SFs in all final answers. }

iii) Calculate the heat needed (in J) in each of the three stages when 7.0 g of ammonia is heated from - 60.0°C to -10.0°C?

a)___ J
b)___ J
c) ___ J

iv) Calculate the total heat released (in kJ) when 85.0 g of diethyl ether is cooled from 53.0°C to 10.5°C. ___ kJ

boiling point

-33.4°C

specific heat capacity, NH  3  (  l  )

4.7 J/(g·  o  C)

heat of vaporization

1380 J/g

specific heat capacity, NH  3  (  g  )

2.2 J/(g·  o  C)

Explanation / Answer

i) In what phase will ammonia be at -60.0oC,

   answer: liquid(L)

ii) In what phase will ammonia be at -10.0oC,

   answer: gas(G)

because, meltingpoint of NH3 = - 77.73 c, boiling point = -33.4°C

iii) ammonia is heated from - 60.0°C to -10.0°C

At, - 60.0°C ammonia is liquid, at -10.0 C ammonia is gas.

heat needed to heat NH3 liquid upto boiling point ,

   = m*s*DT

= 7* 4.7*(-33.4-(-60.0))

= 875.14 j

heat needed to convert NH3 liquid into gas at boiling point ,

    = DHvap*m

    = 1380*7

    = 9660 j

heat needed to heat NH3 gas upto -10.0 c

= m*s*DT

= 7*2.2*(-10.0-(-33.4))

= 360.36 j

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