Chapter 14, Problem 27P

To determine

How much nitrogen evaporates if it is at its boiling point of 72 K and has a latent heat of vaporizing of 200 kJ/kg.

Answer to Problem 27P

Solution:

61.8 g nitrogen evaporates if a 28 g ice cube at its melting point is dropped into an insulated container of liquid nitrogen.

Explanation of Solution

Formula used:

The heat energy absorbed or released to raise the temperature by one degree is given by;

$\begin{array}{l}Q=mc\Delta T\\ Q=mL\end{array}$

Here,

m = mass of the substance.

C = specific heat.

◻T= Temperature difference.

L=latent heat of fusion.

Calculation:

The heat of ice cubes evaporates liquid nitrogen until the temperature of ice cube and liquid nitrogen becomes equal.

Energy released from ice is given by;

$\begin{array}{l}E=mc\Delta T\\ E=\left(273.15\text{ K-77 K}\right)\times 2100\text{ J/kg}\times \text{(28 g}\times \text{1 kg/1000g)}\\ E\text{=12357 J}\end{array}$

The heat of vaporization is 200 kJ/kg or 200x10³ J/kg.

So, vaporization energy is given by;

$H=mL$

Here,

m = mass of liquid nitrogen vaporized.

As energy from the ice goes into H, we get;

$\begin{array}{l}H=mL\\ 12357\text{ J}=m\times 200\times {{\displaystyle 10}}^3\text{ J/kg}\\ m\text{=61}\text{.8 g}\end{array}$