1. A solid of mass 150g of 200 degree celsius is placed in 0.4kg ofwater at 20 degree celsius till a constant temparature is attained. Ifthe specific heat capacity of the solid is 0.5J/g K .Find theresulting temparature of the mixture.2. Calculate the amount of heat energy absorbed by 40g of ice at -10degree celsius to convert into water at 80 degree celsius. Given thatspecific heat capacity of ice is 2.1J/g degree celsius.3. Some hot water is added to three times its mass of cold water at 10degree celsius . The resulting temparature is found to be 20 degreecelsius, find initial temparature of the hot water.4. What mass of a liquid A of specific heat capacity of 0.8J/g K andat temparature of 40 degree celsius must be mixed with 100g of aliquid B of specific heat capacity 2.1J/g K and at 20 degree celsiusso that the final temparature becomes 20 degree celsius.
For water,
From law of conservation of energy,
Heat lost by body = heat gained by water
So,
2) Here, heat absorbed to raise the temperature of ice from
Heat absorbed to raise the temperature of water from
Total heat absorbed = 840 + 336000 = 336840 J