Calorimetry problem.?

Question

Steam at 100'C is passed over 1000g of ice at 0'C.After sometime 600 g of ice at 0'C is left and 450g of water at 0'C is formed.Calculate the specific latent heat of vapourisation of steam.

Given: Specific latent heat of ice is 336J/g and specific heat capacity of water is 4.2 J/g/K

Answer 1

you had 1000g of water. As 600g of ice are left you have melt
1000-600 =400g of ice
For th=his you must give to the ice E=m *specific latent heat ice
400*336=134400 J
As you have 450g of water you can say that this energy was given by cooling 450-400=50g of steam
When you pass from steam at 100C to water at 0C
two steps 1) transform stam in liquid , after 2) cool the liquid from 100 to 0c
the step 2 demands Q= mcdt = 50*(0-100)*4.2 =-21000J

So the transformation needed Q= -mLv -21000
This is energy lost by the steam for this reason, I put the sign -
So you have 134400 = -(-50*Lv -21000)
134400 = 50Lv +21000
Lv = (134400-21000)/50 =2268 J/g

Answer 2

By mass balance, steam condensed
= 600 + 450 - 1000 = 50 g

Heat lost by steam in condensing
= 50L, where L = latent heat of vapourisation of steam (J/g)
Heat lost by hot condensate at 100° to be water at 0°
= 50 * 4.2 * 100 J = 21000 J

Heat gained by 400 g of ice to become water
= 400 * 336 = 134400 J

Heat lost = heat gained
=> 50L + 21000 = 134400
=> L = (134400 - 21000) / (50) = 2268 J/g.