If there are no heat losses , the heat released by the condensation of x gram of steam at `100^@C` into the water at `100^@ C` can be used to convert y gram of ice at `0^@C` into the water at `100^@C` . Then the ratio of y : x is nearly

If there is no heat loss, the heat released by the condensation of x gram of steam at 100^@C into water at 100^@C can be used to convert y gram of ice at 0^@C into water at 100^@C . Then the ratio of y:x is nearly [Given L_l = 80 cal//gm and L_v= 540 cal//gm ]

Heat required to convert 1 g of ice at 0^(@)C into steam at 100 ^(@)C is

Work done in converting 1 g of ice at -10^@C into steam at 100^@C is

How many calories of heat will be required to convert 1 g of ice at 0^(@)C into steam at 100^(@)C

500g of ice at 0°C is mixed with 1g steam at 100°C . The final temperature of the mixture is

Heat released by 1kg steam at 150^(@)C if it convert into 1kg water at 50^(@)C .

Heat required to raise the temperature of one gram of water through 1^@C is

In an energy recycling process, X g of steam at 100^(@)C becomes water at 100^(@)C which converts Y g of ice at 0^(@)C into water at 100^(@)C . The ratio of X/Y will be (specific heat of water ="4200" J kg"^(-1)K, specific latent heat of fusion =3.36xx10^(5)"J kg"^(-2) , specific latent heat of vaporization =22.68xx10^(5)" J kg"^(-1) )

100g ice at 0^(@)C is mixed with 100g water at 100^(@)C . The resultant temperature of the mixture is

100 g ice at 0^(@)C is mixed with 10 g steam at 100^(@)C . Find the final temperature.