Respuesta :
Answer:
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 .
