How much steam at `100^@C` will just melt 64 gm of ice at `-10^@C`?

(a) The temperature of 100 g of water is to be raised form 10^@C to 40^@C by adding steam at 100^@C to it. Determine the mass of the steam required for this purpose. (b) How much steam at 100^@C will just melt 2700 g of ice at -10^@C ?

How much steam at 100^@C will just melt 1600 g of ice at -8^@C ? (Specific heat of ice= 0.5 (cal)/g^@C , specific heat of water= 1/cal/g^@C , latent of fusion of ice = 80 cal/g , and latent heat of vaporisation of water= 540 cal/g )

How much steam at 100°C will just melt 3200g of ice at -10°C ?

How much steam at 100^(@)C is needed to change 40 g of ice at -10^(@)C to water at 20^(@)C if the ice in a 50 g copper can? Assume that the can maintains the same temperature as the ice and water. Given that specific heat of copper is S_(Cu)=0.09 "cal"//gm^(@) and specific heat of ice is S_("ice")=0.5 cal /gm .^(@)C .

10 grams of steam at 100^(@) C is mixed with 50 gm of ice at 0^(@) C then final temperture is

10 grams of steam at 100^(@)C is mixed with 50 gm of ice at 0^(@)C then final temperature is

5 gm of steam at 100^(@)C is passed into six gm of ice at 0^(@)C . If the latent heats of steam and ice in cal per gm are 540 and 80 respectively, then the final temperature is -