Question

Two 50 gram ice cubes are dropped into 200 grams of water in a class. 1.) if the water was initially at a temperature of 25 degrees C, and if the ice came directly from a freezer operating at -15 degrees C, what will be the final temperature of the drink? The specific heat of ice is about 0.50 cal/(gram degree C) in this temperature range, and the heat required to melt ice to water is about 80 cal/gram. 2.) If there is ice remaining, how much is left?

Answer 1

Heat loss by water when it cools down to 0 ^oC, Q_w = 200 * 1 * (25 - 0) = 5000 cal

Heat gain by ice when it heats up to 0 ^oC, Q_ice = (2 * 50) * 0.5 * [0 - (-15)] = 750 cal

So, heat loss by water used in melting ice = Q_w - Q_ice = (5000 - 750) cal = 4250 cal

Let 'x' g of ice is melted by this much heat. Then,

4250 = x * 80

=> x = 53 g < 100 g

So, the final temperatue of the system is 0 ^oC, and the composition of the system is:

Water = (200 + 53) g = 253 g (at 0 ^oC)

Ice = (100 - 53) g = 47 g (at 0 ^oC)