`1kg` of water is at `20^(@)C`. A resistor of `20Omega` is connected across `200V` battery and the heat dissipated is supplied to water then the time taken by the water to evaporate is : (Specific heat `=4200 j//kg ^(@)C`, Latent heat `=2260kj/Kg`)

One kg of water, at 20^(@)C , is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 Omega . The rms voltage in the mains is 200V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to: [specific heat of water= 4200J/kg ""^(@)C ], Latent heat of water = 2260kJ/kgI

A 210 W heater is used to heat 100 g water. The time required to raise the temperature of this water from 25^(@)C to 100^(@)C is (specific heat capacity of water =4200" J/Kg-"^(@)C )

10 litres of hot water at 70^@C is mixed with an equal volume of cold water at 20^@ C . Find the resultant temperature of the water. (Specific heat of water = 4200 J//kg -K ).

10 litres of hot water at 70^(@) C is mixed with an equal volume of cold water at 20^(@) C. Find the resultant temperature of the water. (Specific heat of water = 4200 j/kg K )

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates .Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10kg water and 0.2g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5^(@)C . Specific heat capacity of water = 4200J kg^(-1).^@C^(-1) and latent heat of vaporization of water = 2.27 xx 10^(6)J kg^(-1)

What is the rice in temperature of 5 kg of water if it is given 84,000 J heat energy? Specific heat capacity of water =4200 J kg ^-1^@C^-1 .

Calculate the heat energy required to raise the temperature of 2 kg of water from 10^@ C to 50^@ C. Specific heat capacity of water is 4200 J kg^(-1) K^(-1) .

One kilogram of ice at 0^(@)C is mixed with one kilogram of water at 80^(@)C . The final temperature of the mixture is (Take : specific heat of water =4200 J kg^(-1) K^(-1) , latent heat of ice = 336 kJ//kg^(-1) )