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...

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

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

in this mixing problem

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

heat gained by ice = m*L_ice+m*S_water*(t_f - 0)

                            = 1*80 + 1*1*t_f

Heat lost by steam = m*L_water + m*S_water *(100 - t_f)

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

t_f = 10.5 C

Note: All Units are in cgs

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.