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When 1g of ice ai 0'C is mixed with 1g of steam at 100'C then the temperature of the mixture at the equillibrium is...

6 years ago

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										Dear student,
The temperature of the mixture will be 0C=273k...
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Sagar Singh
B.Tech, IIT Delhi
6 years ago
										

der rohit,


The definition of a calorie is the amount of heat it takes to raise the temperature of
one gram of water by 1°C. But the heat of fusion of ice is 80 calories per gram; that
is, it takes 80 calories just to change 1 gram of ice at 0°C to 1 gram of water at 0°C.
So if you mix the boiling (100°C) water and the 0°C ice together, neglecting any loss
of heat to the outside, the boiling water will give up 80 cal. of heat just to melt the
1 g. of ice to water at 0°C. In doing so, the boiling water reduces its temperature by
80°C (1 calorie per gram per degree) so it's now at 20°C. Now it's like mixing 1 g. of
water at 20° with 1 g. of water at 0° and they will come to equilibrium at 10°C.
The final temperature of the mixture is 10°C


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AMAN BANSAL

6 years ago
										

in this mixing problem


Heat is lost by 100 C steam = Heat gained by 0 C ice


heat gained by ice = m*Lice+m*Swater*(tf - 0)


                            = 1*80 + 1*1*tf


Heat lost by steam = m*Lwater + m*Swater *(100 - tf)


                            = 1*0.5 + 1*1 *(100 - tf)


tf = 10.5 C


Note: All Units are in cgs

4 years ago
										Final Temp of the mix will be 100C.Heat of fusion =80cal/gHeat of vapourization=540cal/gOut of 540, 80 cal is used in fusion of ice ,While remaining 460cal will raise the temp of mix upto 100C .But to go above 100 C , heat should greater than 540 cal.
										
19 days ago
										
1*546+1*(100-)t=1*80+t
546+100-100t=80t
here t is -ve this means the amount of heat released when 1 g of steam condenses is very more enough to convert all the ice into water at 0 degreee c
19 days ago

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