In a thermally isolated environment, you add ice at 0°c and steam at 100°c. (a) determine the amount of steam condensed (in g) and the final temperature (in °c) when the mass of ice and steam added are respectively 79.0 g and 11.2 g. (b) repeat this calculation, when the mass of ice and steam added are interchanged. (enter the amount of steam condensed in g and the final temperature in °c.)

 a) 11.2 g

   b) 3.73 g.

Explanation:

a) If we assume temperature of mixture to be 100°C , heat released by steam will be 11.2 x 540 = 6048 cals and heat gain gained by will be

79 x 80 + 79 x 1 x 100 = 14220 cals . Since former heat is less than later heat ,water will not be warmed up to 100°C. Let equilibrium temperature be t .  

Heat gained by water = 79 x 80 + 79 x 1 x t = 11.2 x 540 + 11.2( 100 - t )

t = 9.4°

amount of steam condensed = 11.2 g.

b) In this case, whole of water will be warmed up to 100°C as steam is much .heat required by water to warm up to boiling point

= 11.2 x 80 + 11.2 x 100 = 2016 cals

amount of steam condensed =  2016 / 540 = 3.73 g .