A student takes a calorimeter of mass `50 g`, made of copper of specific heat `0.1 "cal" g^(-1) C^(@ - 1)`. He puts 100g of water at `40^(@)C` into it. He drops 100g of ice at `- 20^(@)C` into it. He then passes steam so that water starts boiling. Latent heat of fussion is 80 cal/g and of vapourization is `540` cal/g. Specific heat of ice is `0.5 "cal" g^(-1) C^(@ - 1)`
Temperature of water after adding ice into it

Options

A. `0^(@)C`
B. `5^(@)C`
C. `10^(@)C`
D. `20^(@)C`

Answer

A

Explanation

`{:(m_(C) = 50 "grams",(T_(1))_("calorimeter") = 40^(@)C,),(s_(C) = 0.1 "cal" g^(-1).^(@)C,,),(m_(w) = 100 "grams",(m)_(iu) = 100 "grams",),((T_(i))_("water") = 40^(@)C,(T_(i))_("ice") = - 20^(@)C,):}`
Heat required by ice to comes into `0^(@)C`
`= 100 xx 0.5 (20) = 1000` calorie
Heat supplied by water + calorimeter to go upto `0^(@)C`
`= (50 xx 0.1 xx 40) + 100 xx 1 (40)`
`= 200 + 4000 = 4200` cal.
We have 3200 calories remaining which will melt the ice
`(Q)_("available") = (m'_("ice") (L_(F))`
`3290 = (m')_("ice") xx (80) rArr (m'_("ice")) = 40` gram
This shows that temperature of water a addingice will be `0^(@)C`