20 g of steam at 100^(@)C is passed into 100 g of ice at 0^(@)C. Find the resultant temperature if specific latent heat of steam is 540 cal/g., specific latent heat of ice is 80 cal/g and specific heat of water is 1cal//g""^(@)C.

20 g of steam at 100^(@)C is passed into 100 g of ice at 0^(@)C. Find

1.	20 g of steam at 100^(@)C is passed into 100 g of ice at 0^(@)C. Find the resultant temperature if specific latent heat of steam is 540 cal/g., specific latent heat of ice is 80 cal/g and specific heat of water is 1cal//g""^(@)C.
Answer» Solution :LET the resultant temperature of the mixture be `t^(@)C`. In the method of mixtures, heat lost by steam = heat gained by ICE `m_(s)L_(s)+m_(s)S(100-t)=m_(c)L_(c)+m_(c)s(t-0)` where `m_(s)` is the mass of steam, `L_(s)` is the latent heat of steam, s is the specific heat of water, `m_(c)` is the mass of ice and `L_(c)` is the latent heat of ice. Here, `m_(s)=20g,L_(s)=540cal//g,s` = `1cal//g""^(@)C,m_(c)=100G,L_(c)=80cal//g` `20xx540+20xx1(100-t)=100xx80+100xx1xxt` `10800+2000-20t=8000+100t` 120t = 4800 `t=40^(@)C`.