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

10 grams * 2 J/gdegK = 20 J/degK
20J/degK * 10 deg K = 200 Joules to get it all to the boiling point

then boil 5 grams with remaining 200 J

heat/gram = 200 Joules/5 grams
= 40 Joules/gram

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