Question

60g of ice at -300C is heated, until whole of it evaporates. How much heat energy is needed? (Specific heat of ice = 0.5 cal g-1 0 C-1)​

Answer 1

Answer:

Correct option is A)

Step 1:Given data

60g ice at 0∘

60g steam at 100∘C

Latent heat of steam, Lv=540 cal/g

     Latent heat of ice, Lf=80 cal/g

     Specific heat of water =1 cal/g

Step 2 :Heat required 

Heat required to melt  60 g of ice =mLf

=60×80=4800 cal

Heat required to raise the temperature of water at 0∘C=ms△θ

        =60×1×(100−0)

        =6000 cal

Total heat energy required =4800+6000

                                              =10800 cal

Step: Final composition

This energy is given by steam by getting converted into water.

Amount of steam converted into water is :-

Q=mLv

m=LvQ=54010800=20 g

Thus finally, 60+20=80 gm water and 60−20=40 g steam.

Answer 2

Solution,

Mass of ice (Mi) = 60 g

Now

heat required to change - 300 oC ice to 0 oC ice is=

=Mi * Si *(0- (-300))

=60*0.5*300

=9000 cal

heat required to change 0 oC ice to p oC water

= Mi *Lf ( latent heat of fusion

= 60 *80

=4800 cal

heat required to change 0 oC water to 100 oC water

= Mi*Sw*(100-0)

= 60*1*100

=6000 cal

heat required to change 100 oC water to 100 oC vapour

= Mi*Lf ( lf is latent heat of vapourization

=60*540

=32400 cal

total heat required= 9000+4800+6000+32400

= 32400 cal