Question

30g ice cube at 0 degree Celsius is dropped into 200g of water at 30 degree Celsius. Calculate the final temperature of water when all the ice cube has melted. Given: latent heat of ice = 80 cal/g; specific heat capacity of water= 1cal/g

Answer 1

let the final temperature be T..

  T = [ 30 * 1 * 0⁰ + 200 * 1 * 30⁰ -  30 * 80] / (200 + 30)
     = 360 / 23  = 15.6° C

another way:

heat absorbed by ice cube + water: 30 gm * 80 cal/gm + 30gm * 1cal/gm * T °C
 heat given out by water at 30° C:  200 gm * 1 cal/gm * (30° - T)
           =>  2400 + 30 T = 6000 - 200 T
                         T = 3600 / 230 = 15.6°C