The amount of heat required to convert 5g of ice at 0∘C to 5g of steam at 100∘C is - [Latent heat of vaporization and fusion are Lv=540cal g−1 and Lf=80cal g−1]
A
3100cal
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B
3200cal
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C
3600cal
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D
4200cal
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Solution
The correct option is C3600cal Pictorial representation of conversion of ice at (0∘C) into steam at (100∘C) is shown below.
Given, Mass of ice (mi)=5g Temperature of ice (T1)=0∘C Temperature of steam (T2)=100∘C Latent heat of vaporization (Lv)=540cal g−1 Latent heat of fusion (Lf)=80cal g−1
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×(T2−T1)+mw×Lv
∴Q=5×80+5×1×(100−0)+5×540 =3600cal Thus, option (c) is the correct answer.