Physics, asked by suraj367, 1 year ago

20 gram ice mixed with 20 gram water at 60degree celcius find (1) temperature of mixture (2)find out level of water (3) find out ice remain in water

Answers

Answered by arrrjunkhadep4m571
4
Latent heat per gram for ice is 80cal /gram
Specific heat capacity of water is 1cal/g C
For ice it is 0.5 cal/g C

Now,
Latent heat required for 20 grams of ice is mass * latent heat per gram

20* 80 = 1600 calories

The heat transfer between the ice and the water will stop when they reach some equilibrium temperature

If the ice completely melts and the there is still some temperature difference, the heat transfer will continue

But, if it does not melt completely, then the heat transfer once the temperature of the water itself becomes 0 C (no further difference in temperature) and the temperature of the ice will anyway be 0 C as it has not absorbed all the latent heat required

So, the amount of maximum energy that the water can provide before it reaches 0 C is
20 * 1* 60 = 1200 calories

As this is lesser than the latent heat required to melt the entire quantity of ice, we can say that some fraction of the ice has melted and some of it remains in the mixture as ice

Bear in mind that the water at 60 C will become water at 0 C and some amount of the ice will be ice and some of it water

Let the percentage of ice that has become water be x%
Now the maximum of heat by water will be equal to
X/100 * 1600 = 1200

Simplify it and you will get that 75 percent of the ice melts and 25 per cent remains in the mixture And the temperature of the mixture will be 0 C

The total volume of the mixture will be
1/4 * 20 = 5 g ice
Multiplying by its density
You will get that it takes up 4.55 cm3

The total amount of water is 35 g and it's density is 1 g/cm3
And it will take up 35 cm3

Therefore, total volume = 39.55 cm3

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