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 boil

Question

3 years ago

11

ing point?

1 answer:

sladkih [1.3K] · Answer 1 · 2022-07-01T22:41:39+03:00

<h3>Answer:</h3>

= 19.712 kJoules

<h3>Explanation:</h3>

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.