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.
