Question

If 10 g of steam of 100 C is introducded into a mixture of 200 g of water and 120 g of ice find the the final temperature and composition of the mixture?

mass ice= 120g, c=0.5

mass water=200 g , c=1

Lf= 80 cal/ g Lv= 540 cal

Answer 1

Data Given

mass of steam m_s = 10 g, T_s = 100⁰C, L_v = 540 cal, Mass of water m_w = 200 g, c_w = 1 cal/g⁰C mass of ice m_ice = 120 g, L_f = 80 cal/g

Heat Required by ice to melt

Heat lost by steam

Clearly whole the ice will not melt as heat required by ice to melt is greater then the heat lost by steam so final temperature of the mixture will be 0⁰C As well heat available for melting of ice is 6400 cal so the mass of ice which melt is

Now final composition of the mixture will

water available = mass of water + mass of ice melted + mass of steam condensed.

ice at final stage m_ice = 120 g - 80 g = 40 g

So final mixture has 290 g water and 40 g ice at 0⁰ C.