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If 100 g of ice at 0C is mixed with 100 g of boiling water at 100C Graph the variation...
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When 100 g of ice at `0^oC` is mixed with 100 g of boiling water at `100^o C` , heat is transferred from the boiling water to the ice, resulting its melting into water at `0^oC ` and then its temperature will increase gradually till it reaches a thermal equilibrium.
Let the final temperature of the mixture, after thermal equilibrium is established, be `x^o` C.
Amount of heat absorbed in melting the ice=latent heat* mass
`=8000` cal (latent heat of fusion of ice = 80 cal/g)
Amount of heat absorbed in raising the temperature of 100 g water at 0^C to x^oC=mass*specific heat*temperature rise
`=100*1*x` (specific heat of water is 1 cal/g)
Total heat absorbed=`(8000+100x)` cal
This heat will be supplied by the boiling water.
Heat released by its cooling to `x^oC`
By the condition of the equilibrium,
So, final temperature of the mixture is `10^oC`
The graph is shown in the attached figure.
Posted by llltkl on August 23, 2013 at 7:39 AM (Answer #1)
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