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In a container of negligible heat capacity, 200 g ice at 0oC and 100 g steam at 100oC are added to 200g of water that has temperature 55oC. Assume no heat is lost to the surroundings and the pressure in the container is constant 1.0 atm. What is the final temperature of the system?

In a container of negligible heat capacity, 200 g ice at 0oC and 100 g steam at 100oC are added to
200g of water that has temperature 55oC. Assume no heat is lost to the surroundings and the pressure
in the container is constant 1.0 atm. What is the final temperature of the system?
 

Grade:10

1 Answers

Harshit Singh
askIITians Faculty 5963 Points
2 years ago
Dear Student

In a container, 200 gm of ice at 0 degree celsius , 100 gm of steam at 100 degree celsius and 200 gm of water at 55 degree ceelsius are mixed together.

Latent Heat of fusion of ice = 80

Latent heat of vaporization of water = 540

Specific heat capacity of water =

To Find:

Final temperature of the system

Solution:

Heat absorbed = heat rejected.

Heat rejected by steam = mass of steam(latent heat of vaporization)



Heat absorbed by ice = mass of ice (latent heat of fusion)

Heat absorbed by the water which was formed by ice upto 55 degree

Q = mcdT (where c is specific heat of water = 1)

total heat absorbed = 27 Kcal.

Total heat rejected = 54 KCal.

Now we have 300 gm of water at 55 degree and 100gm of water at 100 degree celsius.

Heat absorbed by water at 55 degree celsius upto final temp 100 degree celcius.

Total heat absorbed = 27 + 18 = 45 KCal.

Final temperature of the system = 100 degree celsius.

Regards

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