Question

You fill an ice tray with water at 25 degree celsius and place it into the freezer. Each cube has a volume of 30 mililiters (a mass of 30 g). Assume that heat is transferred from the water in each cube in the ice cube tray to the surrounding cold air in the freezer at a constant rate of 40 J/min. You can consider that the water (whether liquid or solid) in each cube is at uniform temperature. You can consider the heat capacity of the ice tray to be negligible. The pressure equals 0.101Mpa at all times.

A) calculate the temperature of the water/ice after 40 minutes of cooling

B) how many minutes wil it take for each cube to be frozen to a temperature of -10 degrees celsius

Answer 1

Heat removed in 40 min = 40 J/min x 40 min = 1600 J

Heat capacity of water = 4.187 J/g-K

Heat capacity of ice = 2.108 J/g-K

By heat balance,

Heat removed = Sensible heat removed from water till 0 C + latent heat of fusion + sensible heat removed from ice

Now,

Sensible heat removed from water till 0 C = mCp(T-0) = 30 x 4.187 x (25-0) = 3140.25 J > 1600 J

Therefore, till 40 min just the sensible heat is removed and hence the state is liquid.

By heat balance,

1600 = 30 x 4.187 (25-T2)

.: T2 = 12.262 C

B] By heat balance

Q = mCp_water(T1 - 0) + + mCp_ice(0-T2)

.: 40xt = 30x4.187x(25-0) + 334 + 30x2.108x[0-(-10)]

.: t = 102.67 min