Question

A thermos flask of negligible heat capacity contains 100 g of ice and 30 g of water. Calculate :
(i) The mass of steam at 100 °C needed to condense in the flask so as to just melt the ice:
The amount of water in the flask after condensation. (Specific latent heat of vaporization
of steam 2260 Jg-1, Specific latent heat of fusion of ice
336 Jg-1, Specific heat
capacity of water = 4.2 Jg-1 °C).
(iii) Is it possible to condense the water formed, back to ice by adding ice at 0 °C ? Explain
giving a suitable reason to justify your answer.

Answer 1

Explanation:

Let mass of steam passed in the flask be m grams. Then, Heat energy released by the steam, Q = mL

=mgram×2260J/g=2260mJ

Since, Heat given = Heat taken, therefore,

2260m=22600⇒m=

2260

33600

=14.87g

∴ Total amount of water in the flask after condensation =100g+30g+14.87g=144.87g.