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/g
B. 20 J/g
C. 80 J/g
D. 200 J/g
E. 40 J/g
- I am stuck on this!

To find the heat of vaporization of the liquid, we need to first calculate the amount of heat absorbed by the liquid to increase its temperature and then use that information to determine the heat of vaporization.

First, we calculate the heat absorbed by the liquid to increase its temperature. We are given that the specific heat of the liquid is 2 J/g*K, and the liquid increases its temperature by 10 K. So, the heat absorbed can be calculated using the equation:

Heat absorbed = mass * specific heat * temperature change

Given:
Mass = 10 g (sample size)
Specific heat = 2 J/g*K
Temperature change = 10 K

Heat absorbed = 10 g * 2 J/g*K * 10 K
Heat absorbed = 200 J

Now, we need to consider that only half of the liquid boils away before all the heat is used up. Since the heat of vaporization represents the amount of heat required to change one gram of a substance from a liquid to a gas at its boiling point, we can determine the heat of vaporization by dividing the total heat absorbed by the mass of the liquid that boils away. Since half of the liquid boils away, the mass of the liquid that boils away is half of the initial mass (5 g).

Heat of vaporization = total heat absorbed / mass of liquid that boils away

Heat of vaporization = 200 J / 5 g
Heat of vaporization = 40 J/g

Therefore, the heat of vaporization of the liquid is 40 J/g, which corresponds to option E.