Calorimetry - Chemistry

Question

You order a 16 oz glass of tea (where the mass of Tea is 474 grams) from a local restaurant. The tea is freshly brewed and has an initial temperature of 22.84 °C. You add ice to cool it. If the heat of fusion of ice is 6.020 ^kJ/_mol and each ice cube contains exactly 1 mol of water (assume the ice does not add anymore volume), how many ice cubes are necessary to cool the tea to 3.38 °C? The specific heat of the "tea" is 4.184 ^J/_g°C. 

*ice cubes (partial cubes allowed) and ignore the heat required to raise the temperature of the melted ice cubes

Answer 1

How much heat must be transferred from the hot tea to the ice to lower the temperature of the tea from 22.84º to 3.38º?

q = mC∆T

q = heat = ?

m = mass of tea = 474 g

C = specific heat of tea =4.184 J/gº

∆T = change in temperature= 22.84º - 3.38º = 19.46º

q = (474 g)(4.184 J/gº)(19.46º) = 38,593 J of heat must be transferred

Each cube has 1 mol water and to melt one cube requires 6.020 kJ of heat.

So, how many ice cubes will it take to accept the 38,593 J of heat lost by the hot tea?

38,593 J x 1 kJ / 1000 J = 38.593 kJ

38.593 kJ x 1 cube / 6.020 kJ = 6.4 ice cubes