# 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...

Grade:12

## 9 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago
Dear student,
The temperature of the mixture will be 0C=273k...
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Sagar Singh
B.Tech, IIT Delhi
Aman Bansal
592 Points
13 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

Mukesh Singh
33 Points
12 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

Pardeep Kumar
11 Points
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.
sailikith
11 Points
7 years 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
Mahathi
12 Points
6 years ago
100c.because WHEN A STEAM is mixed with the ice the temperature flows from higher to lower
But 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.
peeyush bangar
11 Points
6 years ago
dear rohit
in 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
Gitanjali Rout
184 Points
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)

t= 10.5 C

Note: All Units are in cgs

Gitanjali Rout
184 Points
6 years 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°C

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