Question

Your 300mL cup of coffee is too hot to drink when served at 87.0 ∘C . You wish to cool your coffee to a temperature of 55.0 ∘C which is much more comfortable to drink. To do this, you take an ice cube from a -21.0 ∘C freezer and add it to the cup.

Here is some information that may be helpful:

cice = 2090 Jkg∘C

cwater = 4190 Jkg∘C

Lf = 334,000 Jkg

Lv=22.6×105Jkg

For water 1 g = 1 mL = 1 cm3

How many Q values are required in your calorimetry equation to successfully account for all temperature and phase changes in this problem?

What mass ice cube is required?

Answer 1