Question

Consider a 10 gram sample of a liquid with specific heat 2 J/gK. By the addition of 400 J, the liquid increases its temperature by 10 K and then boils. Half of the liquid boils away before all the heat is used up. What is the heat of vaporization of the liquid?A.10 J/gB.20 J/gC.80 J/gD.200 J/gE.40 J/g© 2018.Grand Canyon University. All Rights Reserved.

Answer 1

40 J/g is the heat of vaporization of the liquid.

Answer: Option D

Explanation:

Given that mass of liquid sample: m = 10 g

And, Specific heat of the liquid: S = 2 J/g K

Also, the increase in the temperature of the liquid,  $\Delta T = T_{2}-T_{1} = 10 K$

Therefore, the total amount of heat energy required is given by:

              $q_{1} = m \times S \times\left(T_{2}-T_{1}\right) = 10 \times 2 \times 10 = 200 J$

According to the given data in the question,

Total heat energy supplied, q = 400 J

Rest of heat would be $q_{2}=q-q_{1}=400-200=200 \mathrm{J}$

Now, 200 J vaporizes the mass, half of the liquid from full portion boiled away. So,

                 $m^{\prime} = \frac{10}{2} = 5 \mathrm{g}$

Latent heat of vaporization of the liquid is $L_{v}$. It can be calculated as below,

                       $q_{2} = m^{\prime} L_{v}$

                       $L_{v} = \frac{q_{2}}{m^{\prime}} = \frac{200}{5} = 40 \mathrm{J} / \mathrm{g}$