Water at `80^o` C is mixed with water at `40^o` C to get 20kg water at `50^o` C. the mass of the hot water taken is

If hot water is mixed with cold water

15 g ice at 0^@C is mixed with 10 g water at 40^@C . Find the temperature of mixture. Also, find mass of water and ice in the mixture.

0.1 m^(3) of water at 80^(@)C is mixed with 0.3m^(3) of water at 60^(@)C . The finial temparature of the mixture is

10 g of water at 70^@C is mixed with 5 g of water at 30^@C . Find the temperature of the mixture in equilibrium. Specific heat of water is 1 cal//g-^@C .

10 g of water at 70^@C is mixed with 5 g of water at 30^@C . Find the temperature of the mixture in equilibrium. Specific heat of water is 1 cal//g.^@C .

A tap supplies water at 15^@C and another tap connected to geyser supplies water at 95^@C . How much hot water must be taken so as to get 60 kg of water at 35^@C ?

80 g of water at 30^(@) C is mixed with 50 g of water at 60^(@) C , final temperature of mixture will be

200 g of hot water at 80^@ C is added to 300 g of cold water at 10^@ C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg^(-1)K^(-1)

If 10 g of ice at 0^(@)C is mixed with 10 g of water at 40^(@)C . The final mass of water in mixture is (Latent heat of fusion of ice = 80 cel/g, specific heat of water =1 cal/g""^(@)C )

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)