How many gm of ice at `-20^(@)C` are needed to cool 200 gm of water from `25^(@)C` to `10^(@)C`?

How many grams of ice at -14 .^(@)C are needed to cool 200 gram of water form 25 .^(@)C to 10 .^(@)C ? Take specific heat of ice =0.5 cal g^(-1) .^(@)C^(-1) and latant heat of ice = 80 cal g^(-1) .

How many gram of ice at -14^@C are needed to cool 200 gm of water from 25^@C to 10^@C . Specific heat of ice = 0.5 cal/gm ^@C and latent heat of ice = 80 cal/gm.

What amount of ice at -14^(@)C required to cool 200gg of water from 25^(@)C to 10CC? (Given C_("ice")=0.5cal/g.^(@)C,L_(f) for ice =80cal//g )

How many grams of ice at 0^@C should be added to 200 gm. of coffee at 90^@C to cool it to 60^@C ? (Assume that coffee-cup is not heated) specific heat of water = 1 c a l / g m ∘ C , Specific heat of coffee = 1.4 c a l / g m ∘ C

10 gm of ice at -20^(@)C is dropped into a calorimeter containing 10 gm of water at 10^(@)C , the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

10 gm of ice at -20^(@)C is dropped into a calorimeter containing 10 gm of water at 10^(@)C , the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

10 gm of ice at -20^(@)C is dropped into a calorimeter containing 10 gm of water at 10^(@)C , the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

10 gm of ice at – 20^(@)C is added to 10 gm of water at 50^(@)C . Specific heat of water = 1 cal//g–.^(@)C , specific heat of ice = 0.5 cal//gm-.^(@)C . Latent heat of ice = 80 cal/gm. Then resulting temperature is -