`m_(1)` gram of ice at `-10^(@)C` and `m_(2)` gram of water at `50^(@)C` are mixed in insulated container. If in equilibrium state we get only water at `0^(@)C` then latent heat of ice is :
M gram of ice at 0^@ C is mixed with 3 M gram of water at 80^@ C then the final temperature is.
50 gram of ice at 0^(@) C is mixed with 50 gram of water at 60^(@)C , final temperature of mixture will be :-
An insulated container has 60 g of ice at -10^(@)C . 10 g steam at 100^(@)C , sourced from a boiler, is mixed to the ice inside the container. When thermal equilibrium was attained, the entire content of the container was liquid water at 0^(@)C . Calculate the percentage of steam (in terms of mass) that was condensed before it was fed to the container of ice. Specific heat and latent heat values are S_("ice") = 0.5 cal g^(-1) .^(@)C^(-1) , S_("water") = 1.0 cal g^(-1) .^(@)C^(-1) L_("fusion") = 80 cal g^(-1) , L_("vaporization") = 540 cal g^(-1)
Find the final temeprature and composition of the mixture of 1 kg of ice at -10^(@)C and 4.4 kg of water at 30^(@)C . Given that specific heat of water is 4200J//kg^(@)C and that of ice is 2100J//kg^(@)C and latent heat of fusion of ice is 3.36xx10^(5) J/kg
M g of ice at 0^@C is mixed with M g of water at 10^@ c . The final temperature is
1 kg of ice at 0^(@)C is mixed with 1.5 kg of water at 45^(@)C [latent heat of fusion = 80 cal//gl. Then
1 g of steam at 100^@C and an equal mass of ice at 0^@C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 cal//g , latent heat of ice =80 cal//g ,specific heat of water =1 cal//g^@C )
10 g of ice at 0^@C is mixed with 100 g of water at 50^@C . What is the resultant temperature of mixture
One gram of ice at 0^(@)C is added to 5 gram of water at 10^(@)C . If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -
