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.

1) For the body,
                           m1=150 g=0.15 kgΔT1=200ο-T0s1=0.5 J g-1K-1=500 J kg-1K-1T0=?
For water,
                 m2=0.4 kgΔT2=T0-20οs2=4200 J kg-1K-1T0=?

From law of conservation of energy,

Heat lost by body = heat gained by water

So,        m1s1ΔT1 =m2s2ΔT2

 0.15×500 (200-T0) =0.4×4200 (T0-20)75  (200-T0) =1680 (T0-20)15000-75 T0 =1680 T0-336001605 T0=18600or, T0=186001605=11.5οC

2) Here, heat absorbed to raise the temperature of ice from -10οC to 0οC

                 ΔQ1=s m ΔT         =2100×0.04×10=840 J

Heat absorbed to raise the temperature of water from 0οC to 80οC

               ΔQ2=s m ΔT         = 4200×1×80=336000 J

Total heat absorbed = 840 + 336000 = 336840 J
 

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