Question

100 gram ice at 0°C is mixed with 20 gram steam
at 100°C. Assuming no heat loss to the
surrounding, the final temperature of the mixture is

Answer 1

Explanation:

heat required to take 100gram ice to 100°C water is

= 100×80 + 100×100

= 18000 calories

now heat required to take 20gram steam to 100°C water is

= -20×540

= -10800

Here -ve shows that heat has to be taken out of the system.

now net heat given is 7200 calories so this heat will be taken out of the system to bring out equilibrium ...

so 7200 = 120x

here x is temperature fallen to bring equilibrium.

so x = 60°C

so the temperature will fall 60°C from 100°C

hence final temperature is 40°C.

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