Question

Latent heat of ice is 80 cal per g. Express it in joule per kg. 1 cal is equal 4.18 J

Answer 1

Answer:

=3.34×10^5 J/Kg

Explanation:

=80×4.2

=334 KJ/Kg

=334×10^3

=3.34×10^5 J/Kg

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Answer 2

Answer:

Latent heat of ice = 80 cal/g

Given, 1cal = 4.18 J

So latent heat in J/g

=> (80*4.18) J/g

=> 334.4 J/g

We require it's value in J/kg

1 gram = 334.4

=> 1 Kg = 1000g = 1000*334.4

=> 334400 J/Kg