A thermos-flask of negligible heat capacity contains 100g of ice and 30g of water. Calculate
(i)  the mass of steam of 100$°$C needed to condense in flask so as to just melt the ice.
(ii)  The amount of water in the flask after condenstion. (Latent heat of ice = 336 J g^-1. Latent heat of steam = 2260 J g^-1. Sp: heat capacity of water = 4.2 J g^-1 $°$C^-1)
(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.

Dear student
Please find below the solution to the asked query:

a) the amount of heat that would be lost by the steam while condensing  will be equal to the amount of  heat that would be gained by the ice while melting 

total heat lost in condensation+ heat lost in coming to the temperature of the flask = total heat gained in melting

m_steamL +mct = m_icel

m*2260 + m*4.2*100= 336 * 100

m = 12.54 gm
Thus 12.54g of steam must condense in order for the ice to melt 

 
b) total mass of water in the flask = mass of water from melting ice + initially present mass of water + mass of water formed from the condensation of steam =  100+30+52.01= 182.01 gm


c) no it,s not possible to condense the water formed back to ice by adding ice at 0 degree Celsius because the latent heat required to be removed for forming ice would be given to ice added and the added ice will melt gaining latent heat and will melt and end result will be a equilibrium .
 
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