Question

Calculate the heat absorbed in K cal to convert 250 gm of water at 100 °C into steam at 100 °C​

Answer 1

Answer:

if there is no heat loss, the heat released by the condensation of x gram of steam at `100^@C` into water at `100^@C` can be used to convert y gram of ice at `0^@C` into water at `100^@C`. Then the ratio of y:x is nearly [Given `L_l = 80 cal//gm` and `L_v= 540 cal//gm`]

Explanation:

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