Question

The heat of vaporization for liquid zinc is 1.76 kj/g. How much heat is needed to boil 11.2 g of liquid zinc already at its boiling point?

Answer 1

Answer:

= 19.712 kJoules

Explanation:

• Heat of vaporization refers to the amount of heat required to change a unit mass of a substance from liquid to gaseous state without change in temperature.

To calculate the amount of heat, we use,

Amount of heat = Mass × Heat of vaporization

Q = m×Lv

Given;

Mass of liquid Zinc = 11.2 g

Lv of liquid Zinc = 1.76 kJ/g

Therefore;

Q = 11.2 g × 1.76 kJ/g

  = 19.712 kJ

Thus, the amount of heat needed to boil 11.2 g of zinc is 19.712 kilo-joules.