Question

86
60 g of ice at 0°C is mixed with 60 g of
steam at 100°C. At thermal equili-
brium, the mixture contains (Latent
and 80 cal g
heats of steam and ice are 540 cal g!
heat of water = 1 cal g 1 °C-1)
(1) 80 g of water and 40 g of steam at
1
100°C
(2) 120 g of water at 90°C
(3) 120 g of water at 100°C
(4) 40 g of steam and 80 g of water at
0°C​

Answer 1

Explanation:

40 g of steam and 80 g of water at

0°C

is correct answer please mark as best answer

Answer 2

Explanation:

40 g of steam and 80 g of water at

0°C