Two 45 g ice cubes are dropped into 390 g of water in a ther

Two 45 g ice cubes are dropped into 390 g of water in a thermally insulated container. If the water is initially at 15degreeC, and the ice comes directly from a freezer at - 19degreeC, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? The specific heat of water is 4186 J/kg.K. The specific heat of ice is 2220 J/kg.K. The latent of fusion is 333 kJ/kg.

Solution

a)

Heat lost by water = 0.39*4186*(15 - T) <------ T = final temperature at equilibrium

Heat gained by ice cubes = 2*0.045*(2220)*19 + m*333000 <---- m = mass of ice melting to water

So, by energy conservation,

0.39*4186*(15 ) = 2*0.045*(2220)*19 + m*333000 <----- assuming the final temperature to be 0 deg C

So, m = 0.0621 kg = 62.1 g

So, final temperature will be 0 deg C <--------answer

b)

If one ice cube is used:

0.39*4186*(15 -T ) = 1*0.045*(2220)*19 + 0.045*333000 + 0.045*4186*T <----- assuming all the 45 g of ice melts to water

So, T = 4.2 deg C