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In a container of negligible heat capaci...

In a container of negligible heat capacity, 200 g ice at `0^(@)C` and 100 g steam at `0^(@)C` are added to 200 g of water that has temperature `55^(@)C` Assume no heat is lost to the surroundings and the pressure in the container is constant 1.0 atm.
At the final temperature, mass of the total water present in the system, is

A

`472.6g`

B

`483.3g`

C

`493.6g`

D

`500g`

Text Solution

Verified by Experts

The correct Answer is:
B
As steam has comparatively large amount of heat to provide in the form of latent heat we check what amount of heat is required by the water and ice to go up to `100^(@)C` , that is
`(m_(1)L+m_(i)S_(w) DeltaT)+m_(w).S_(w).DeltaT`
`=[(200xx8)+(200xx1xx100)]+(200xx1xx45)`
=45,000 cal
That is given by m mass of steam, then
`m_(s)L=45,000`
`m_(s)=(45,000)/(540)=(500)/(6)=83.3g`
therefore 83.3 gm steam converts into water of `100^(@)C` . Total water
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