# Temperature after mixing ice and water

## Homework Statement

5 g of water at 30°C and 5 g of ice at -20°C are mixed together in a calorimeter Find the final temperature of the mixture. Water equivalent of the calorimeter is negligible,specific heat of ice=0.5 cal/g°C and latent heat of ice =80 cal/g.

## Homework Equations

C= Q/mΔT
L=Q/m where L-----> Latent heat

## The Attempt at a Solution

The heat lost by the water is gained by the ice.But I don't really know what to do with the formula without knowing the temp change.
The Heat req. to melt the ice is Q=Lm= 400 cal. So should I equate this to CmΔT of water?
Please give me a hint as to how to proceed with this problem. And also I don't understand what the "Water equivalent of the calorimeter is negligible" is supposed to imply.

Last edited:

The Heat req. to melt the ice is Q=Lm= 400 cal. So should I equate this to CmΔT of water?
Please give me a hint as to how to proceed with this problem. And also I don't understand what the 'Water equivalent of the calorimeter is negligible' is supposed to imply.
by water equivalent of a calorimeter ,one knows how much heat will be taken by the container to raise its temperature by one degree equivalent to amount of water
being given the same heat to raise its temp by one degree..

For any given substance the equivalent mass of water W having the same specific heat as the given substance is called its Water Equivalent.
if your calorimeter does not take a lot of heat 'it can be neglected'

now to deal with some events

posed above - one should apply common sense- suppose you take some water at room temp and mix ice at 0 degree and go on adding ice ;
finally you get a mixture of water and ice at temp. 0 degree centigrade.
you have here ice at -20 so the water will provide heat energy to first bring the ice to zero degree then cam melt some ice - now the final temp can be taken to be say Tdegree of the mixture and
heat lost and heat gained can be equated .

SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

5 g of water at 30°C and 5 g of ice at -20°C are mixed together in a calorimeter Find the final temperature of the mixture. Water equivalent of the calorimeter is negligible,specific heat of ice=0.5 cal/g°C and latent heat of ice =80 cal/g.

## Homework Equations

C= Q/mΔT
L=Q/m where L-----> Latent heat

## The Attempt at a Solution

The heat lost by the water is gained by the ice.But I don't really know what to do with the formula without knowing the temp change.
The Heat req. to melt the ice is Q=Lm= 400 cal. So should I equate this to CmΔT of water?
Please give me a hint as to how to proceed with this problem. And also I don't understand what the "Water equivalent of the calorimeter is negligible" is supposed to imply.
To summarize what drvrm said, when you mix ice and water, the water gets colder and the ice gets hotter. Or, did you expect a different result?

Calculate how much heat it takes to bring the temperature of the ice to its melting point and see if the water can supply that much heat.

Analyze what happens in stages.

I understand now. So I divide into two stages and then check.
1) Raise temp of ice to melting point
Q=CmΔT=0.5*20*5 =50J
2) Melt the ice
Q=Lm =80*5 =400J
Total heat required = 450J
3) Check if water can provide this heat
Q=CmΔT= 1*5*30 =150J.
It cannot.So only some of the ice will melt?
m=Q/L =100/80=1.25 g of ice melts. And the temp sticks to 0°C.

If I was given say 6g of water instead, I would be able to release 180J. So that means 30 J will go into further raising the temp of the melted ice form 0 to 3°C? (that is adding the total mass to be 10g of water) .Is that correct?

SteamKing
Staff Emeritus
Homework Helper
I understand now. So I divide into two stages and then check.
1) Raise temp of ice to melting point
Q=CmΔT=0.5*20*5 =50J
2) Melt the ice
Q=Lm =80*5 =400J
Total heat required = 450J
3) Check if water can provide this heat
Q=CmΔT= 1*5*30 =150J.
It cannot.So only some of the ice will melt?
m=Q/L =100/80=1.25 g of ice melts. And the temp sticks to 0°C.
Look at it this way:

The heat exchanged from the water to raise the temperature of the ice to 0° C is 50 J. The ice is at 0° C, but what is the temperature of the water, now that it has lost 50 J of heat to the ice?

When some of the ice starts to melt, you will be mixing the melt water at 0°C with the 5 g of slightly warmer water, current temperature as yet unknown. The 100 J which you think is available to melt ice must also be used to raise the temperature of the mixture of the melt water and the original warmer water to some equilibrium temperature, where all of the liquid water has this same temperature. Therefore, 100 J of heat cannot be assumed to go into totally into melting part of the ice.

If I was given say 6g of water instead, I would be able to release 180J. So that means 30 J will go into further raising the temp of the melted ice form 0 to 3°C? (that is adding the total mass to be 10g of water) .Is that correct?