Question

5 grams of ice at 0 degree C.is mixed with 10 grams of water at 50 degrees C.then resultant temperature is

Answer 1

latent heat of fusion of ice = 336 J/g
specific heat of water = 4.186 J/g degreeC
First 5g ice melts and then its temperature increases by gaining heat from 10g water. let the resultant temperature is T.

Energy gained by 5g ice = energy lost by 10g water
5*(336) + 5*4.186*(T-0) = 10*4.186*(50-T)
1680 + 20.93T = 2093 - 41.86T
(20.93+41.86)T = 2093-1680
62.79T = 413
T = 413/62.79 = 6.58 degree celsius

Answer 2

Resultant temperature is T.

Ice could melt and become water. Water at 50° C will cool down.

Heat lost by Hot water = Heat gained by ice
         = heat energy for melting of 5g of ice
             + heat energy for increase of temperature of 5g water from 0°C to T°C

           m1 s1 (T2 - T)  =  m2 * L + m2 s2 (T - T1)
           
    10g * 4.181J/g/°K * (50 - T) °K = 5g * 334 J/g + 5g * 4.181J/g/°K * (T - 0) °K
  
 or,        10 * 4.181 (50-T) = 5*334 + 5 * 4.181 (T-0) 

          41.81 (50-T) =  1,670 + 20.905 T

             T =  420.5 / 62.715 = 6.705° C