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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...

8 years ago

Dear student,

The temperature of the mixture will be 0C=273k...

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

8 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°CWe are all IITians and here to help you in your IIT JEE preparation.

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.

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AMAN BANSAL

8 years ago

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 CNote: All Units are in cgs

7 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.

2 years 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

2 years ago

100c.because WHEN A STEAM is mixed with the ice the temperature flows from higher to lowerBut it is not much lower as 0c . But the whole material come to an equilibrium state so the ice is converted into water at 100c.thus attains equilibrium.

2 years ago

dear rohitin such questions when we nedd to mix 2 states such as ice ans water at different we need a reference point for mixing them here let us take our refernce point as water at 100celsius. now for ice to convert in water 100celsius energy required will be 180 calories(apply latent and specific heats for conversion to ice to water and water to 100celsius water).same with steam and energy given by steam will be 540 calories.now we can say that ice requires less heat for temp conversion so ice will whole convert to water at 100celsius. but still sone energy is left so it will convert to steam b y calculation we find 2/3 grams of water will convert to steam at 100 celsius and rest will remain water at 100 celsius so temp is 100celsius

one year ago

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 CNote: All Units are in cgs

one year ago

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°CWe are all IITians and here to help you in your IIT JEE preparation.

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.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar : Click hereto download the toolbar..

one year ago

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