Convert into grades: 25° 46′ 6
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Step-by-step explanation:
To convert 20 gm of ice from 20 gm of water at 0∘ celsius,
Latent heat of fusion is 80 Cal/gm
So, for 20 gm, the heat energy required is( Q0 ) = 80*20 = 1600 Cal.
To convert 20 gm of water at 0∘ C to 20 gm of water at 100∘ C,
Q1=20∗1∗(100–0) = 2000 Cal.
To convert 20 gm of water at 100∘ C to steam,
Latent hear of vaporization of water is 540 Cal/gm
Heat requires, Q2=540∗20=10,800 Cal
So, Total heat required = Q0+Q1+Q2=14,400 Cal
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