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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...Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.All the best.Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..Askiitians ExpertSagar SinghB.Tech, IIT Delhi
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6 years ago
```										der rohit,
The definition of a calorie is the amount of heat it takes to raise the temperature ofone gram of water by 1°C. But the heat of fusion of ice is 80 calories per gram; thatis, 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 lossof heat to the outside, the boiling water will give up 80 cal. of heat just to melt the1 g. of ice to water at 0°C. In doing so, the boiling water reduces its temperature by80°C (1 calorie per gram per degree) so it's now at 20°C. Now it's like mixing 1 g. ofwater 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
Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
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+t546+100-100t=80there 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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