Question

100 gm of ice at 0°C is mixed with 100 gm of water at 40°C.Calculate the final temperature and the masses of ice and water.

Answer 1

hey mate!!

here is your answer... .

we have mixed 100 grams of ice at 0°c with 100 grams of water at 40°c

given,

latent heat of fusion = 40°c
specific heat of water = 1 Cal
specific heat of ice = 1/2 Cal

540 g of ice at 0°C is mixed with 540 g of water at 80°C. What is the final temperature of the mixture?

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11 ANSWERS



Atul Mittal, Assistant Engineer at Madhya Pradesh Power Generating Company ,Jabalpur (2014-present)

Updated Wed

Thanks for A2A


We have mixed 540 g of ice at 0°c with 540 g of water at 80°c. 

Given that 
latent heat of fusion = 80 cal/g 
sp. heat capacity of water =1 cal/g/°c 
sp. heat capacity of ice =0.5 cal/g/°c 

So first thing in this mixing process will occur, is heat transfer from water at 40°c to saturated ice at 0°c till thermodynamic equilibrium occurs.
Since ice is saturated, so it will absorb heat in the form of latent heat to convert into water.
Let final temp.( equilibrium) of mixture is = T°c
So energy balance equation,

Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 80°c to final temp T°c

=> (100*40) + {100*1*(T-0)} = {100*1*(40-T)}

=> 100*T = -100*T

=> 200*T = 0

=> T = 0

hope it helps!

Answer 2

The answer is peovided in the attachment.

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