Question

Change into degrees: 18g 10′ 40″

Answer 1

Answer:

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