One kilogram of ice at `0^(@)C` is mixed with one kilogram of water at `80^(@)C`. The final temperature of the mixture is (Take : specific heat of water `=4200 J kg^(-1) K^(-1)`, latent heat of ice `= 336 kJ//kg^(-1)`)
336 g of ice at 0^@C is mixed with 336 g of water at 80^@C . What is the final temperature of the mixture?
One kg of ice at 0^(@)C is mixed with 1 kg of water at 10^(@)C . The resulting temperature will be
If 10 g of ice is added to 40 g of water at 15^(@)C , then the temperature of the mixture is (specific heat of water = 4.2 xx 10^(3) j kg^(-1) K^(-1) , Latent heat of fusion of ice = 3.36 xx 10^(5) j kg^(-1) )
1kg ice at -10^(@) is mixed with 1kg water at 50^(@)C . The final equilibrium temperature and mixture content.
1kg ice at -10^(@)C is mixed with 1kg water at 100^(@)C . Then final equilirium temperature and mixture content.
50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)
LAKHMIR SINGH & MANJIT KAUR-MODEL TEST PAPER 2-SECTION B
