Question

A mass m = 0.043 kg of benzene vapor (L_v = 3.94x10⁵ J/kg) at its boiling point of 80.1°C is to be condensed by mixing with water at 32.0°C. What is the minimum mass of water required to condense all of the benzene vapor? Assume the mixing and condensation take place is a perfectly insulating container.

Answer 1

Let us determine the minimal amount of energy needed to be taken away from the benzene to condensate it:

$Q_- = L_v \cdot m = 3.94\cdot10^5\,\mathrm{J/kg}\cdot 0.043\,\mathrm{kg} \approx 1.7\cdot10^4\,\mathrm{J}.$ If we take away this amount, all the benzene will condensate and it will have the same temperature of 80.1°C.

Now we'll determine the amount of water, that can heat from 32.0°C to 80.1°C due to obtained heat $Q_-$ :

$m_w c_w \Delta T \Rightarrow m_w = \dfrac{Q_-}{c_w\Delta T} = \dfrac{1.7\cdot10^4\,\mathrm{J}}{4.2\cdot10^3\,\mathrm{J/kg/K}\cdot48.1\,\mathrm{K}} = 0.084\,\mathrm{kg}.$

If we take such an amount of water, the water will heat up to 80.1°C and the benzene will condensate. So the temperature of mixture will be 80.1°C.

If we take smaller amount of water, it will heat up to 80.1°C, but not all the benzene will condensate.

If we take larger amount, he benzene will condensate and cool.