Question

A cube of ice of mass 50 gram at zero degree celsius is dropped into 200 gram of water at 50 degree celsius calculate the final temperature of the mixture

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

Answer 2

Let t be the final temperature.

Then heat Liberated by water = m x c x Δt

= [200 x 10-3 x 4.2 x 103 x (50 – t)]

= (42,000 – 840t)

Heat absorbed by ice to change into water at 0oC = 40 x 10 x 336 x 103 = 13,440 J.

Heat absorbed by water to change its temperature from 0oC to toC = (40 x 10-3 x 4.2x 103 x t) = 168t

Total heat absorbed by water = (13,440 + 168t) J

According to the principle of calorimetry,

Heat given = Heat taken

or 42000 – 840t = 13440 + 168t

or 42000 – 13440 = 168t + 840t

or 1008t = 28560

or t = 28560/1008 = 28.33oC

Hence, the final temperature of water is 28.33oC