Thermal properties of matter

Question

If one gram of steam is at 100 degree celsius is mixed with the one gram of ice at 0° C then result and temperature of the mixture is

Answer 1

Let the mixture become water at T°C at steady state.

To convert steam at 100°C to water at T°C

 Q = 1 x 540 + 1 x 1 x (100−T)

Similarly, to convert 1 g of ice to water at T°C

Q = 1 x 80 + 1 x 1 x (T − 0)

Since, both these heats should be the same,

540 + 100 − T = 80 + T

⇒ 2T = 560

⇒ T = 280°C

Since, this is not a possible value of T hence water wont exist and it will be precisely steam at 100°C