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Question

The amount of heat required to convert 5 g of ice at 0C to 5 g of steam at 100C is -
[Latent heat of vaporization and fusion are Lv=540 cal g1 and Lf=80 cal g1]

A
3100 cal
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B
3200 cal
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C
3600 cal
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D
4200 cal
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Solution

The correct option is C 3600 cal
Pictorial representation of conversion of ice at (0C) into steam at (100C) is shown below.


Given,
Mass of ice (mi)=5 g
Temperature of ice (T1)=0C
Temperature of steam (T2)=100C
Latent heat of vaporization (Lv)=540 cal g1
Latent heat of fusion (Lf)=80 cal g1

Since there are no other sources and sinks which can change the amount of substance used in the problem, we can say that mass of steam = mass of water = initial mass of the ice.
i.e mi=mw=ms
Total heat required during the phase change is
Q=Q1+Q2+Q3
Q=mi×Lf+mw×cw×(T2T1)+mw×Lv

Q=5×80+5×1×(1000)+5×540
=3600 cal
Thus, option (c) is the correct answer.

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