# Ch13 Sample Problem #23

```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&deg;C to its boiling point of –33.3&deg;C, allow
it to evaporate, and then continue warming to 0.0&deg;C, how much energy must you
supply?
Step 1: Figure out how much heat is needed to warm the liquid to its boiling point.
qliquid = (12 kg x 1000 g/1 kg)(4.7 J/g•K)(–33.3&deg;C – –50.0&deg;C)
= 941,880 J x 1 kJ/1000 J = 9.4188 x 102 kJ – &gt; 9.4 x 102 kJ (sig. digits)
Step 2: Figure out how much heat is needed to evaporate the liquid to its boiling point.
qevaporation = (12 kg x 1000 g/1 kg)(1 mol/17.0 g)(23.33 kJ/mol)
= 16468 kJ – &gt; 1.6 x 104 kJ (sig. digits)
Step 3: Figure out how much heat is needed to warm the gas.
qevaporation = (12 kg x 1000 g/1 kg)(2.2 J/g•K)(0.0&deg;C – –33.3&deg;C)
= 879120 J x 1 kJ/1000 J = 8.7912 x 103 kJ – &gt; 8.8 x 102 kJ (sig. digits)
Step 4: Add all the heats together
qtotal = 9.4 x 102 kJ + 1.6 x 104 kJ + 8.8 x 102 kJ = 1.782 x 104 kJ
= 1.8 x 104 kJ
```