Question

The latent heat of vaporization for 1 gram water is 535 calorie its value in joule per kg will be

Answer 1

Answer:

Since, 1 calorie per gram = 4186.799993 joule per kilogram.

So, 536 calories per gram is expressed as 2244124.7965 joule per kilogram.

Answer 2

Answer:

latent heat of steam in Joule per Kg is 2.2512 × 10^6 J/Kg.

latent heat of steam = 536 cal/gm

calorie is a unit of energy. 1 calorie is generally equal to 4.2 Joule.

i.e., 1 cal = 4.2 Joule

and obviously you know 1kg = 1000gram

so, 1gram = 10^-3 kg

now let's convert cal/g into J/kg !

latent heat of steam = 536 cal/gm

= 536 (4.2 Joule )/(10^-3 kg)

= 536 × 4.2 × 1000 J/kg

= 536 × 4200 J/kg

= 2,251,200 J/kg

= 2.2512 × 10^6 J/Kg

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