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A. 1720 kJ
B. 125.6 kJ
C. 3440 kJ
D. 4730 kJ

A. 1720 kJ B. 125.6 kJ C. 3440 kJ D. 4730 kJ-example-1
User Tieru
by
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1 Answer

3 votes

Answer:

Q = 3440Kj

Explanation:

Given data:

Mass of gold = 2kg

Latent heat of vaporization = 1720 Kj/Kg

Energy required to vaporize 2kg gold = ?

Solution:

Equation

Q= mLvap

It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg

by putting values,

Q= 2kg × 1720 Kj/Kg

Q = 3440Kj

User Anirban Hazarika
by
7.9k points

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