Question

You have 400.0 grams of room temperature water at 20.0 degrees Celsius in a thermos flask. You drop 130.0 grams of ice at 0.00 degrees Celsius into the thermos and shut the lid.
(a) What is the equilibrium temperature of the system?
(b) How much ice is left in grams? Answer up to 3 significant figures.

Answer 1

Temperature of water (400.0g) = 20.0℃

Mass of ice dropped = 130.0g at 0.00℃

Latent heat of fusion of ice = 333.55J/g

Specific heat of water = 4.184J/g.℃

Let x grams of ice left in the thermos , ice melted =( 130.0-x)g

The final temperature in the thermos should be 0.00℃ if there is ice left in the thermos.

Amount of heat used to melt ice = 333.55J/g × (130.0-x)g

Amount of heat obtained from decrease of temperature from 20.0℃ to 0.00℃ = m.s.∆T = 400.0g × 4.184J/g.℃ ×20.0℃

Heat lost from water is equal to gain by ice to melt.

Hence 400×4.184×20.0 = 333.55×(130.0-x)

33472 = 43361.5 - 333.55x

x = 9889.5/333.55 = 29.649g

(a) The equilibrium temperature of the system is 0.00 ℃ (Answer)

(b) mass of ice left = 29.649g = 29.6 g. (Answer)