The power of a system which can convert `10 kg` of water at `30^@ C` into ice `0^@ C` in one minute `(L_(ice) = 336000 J//Kg` and `S_(water) = 4200 J//kg//K`) will be

Find the quantity of heat required to convert 40 g of ice at -20^(@) C into water at 20^(@) C. Given L_(ice) = 0.336 xx 10^(6) J/kg. Specific heat of ice = 2100 J/kg-K Specific heat of water = 4200 J/kg-K

Find the quantity of heat required to convert 40 g of ice at -20^(@) C into water at 20^(@) C. Given L_(ice) = 0.336 xx 10^(6) J/kg. Specific heat of ice = 2100 J/kg-K Specific heat of water = 4200 J/kg-K

Find the quantity of heat required to convert 40 g of ice at -20^(@) C into water at 20^(@) C. Given L_(ice) = 0.336 xx 10^(6) J/kg. Specific heat of ice = 2100 J/kg-K Specific heat of water = 4200 J/kg-K

Find the result of mixing 0.5 kg ice at 0^@C with 2 kg water at 30^@C . Given that latent heat of ice is L=3.36xx10^5 J//kg and specific heat of water is 4200 J//kg//K .

Find the result of mixing 0.5 kg ince at 0^@C with 2 kg water at 30^@C . Given that latent heat of ice is L=3.36xx10^5 J//kg and specific heat of water is 4200 J//kg//K .

When a small ice crystal is placed in overcooled water it begins to freeze instantaneously. i. What amount of ice is formed from 1 kg of water over cooled to -8^@C ? L of water =336xx10^3 J//kg and s of water =4200 J//kg//K . ii. What should be the temperature of the overcooled water in order that all of it be converted into ice at 0^@C ?

When a small ice crystal is placed in overcooled water it begins to freeze instantaneously. i. What amount of ice is formed from 1 kg of water over cooled to -8^@C ? L of water =336xx10^3 J//kg and s of water =4200 J//kg//K . ii. What should be the temperature of the overcooled water in order that all of it be converted into ice at 0^@C ?

Calculate the amount of heat given out while 400 g of water at 30^@C is cooled and converted into ice at -2^@C . Specific heat capacity of water = 4200 J/kg K Specific heat capacity of ice = 2100 J/ kg K Specific latent heat of fusion of ice = 33600 J/kg