Question

Your 300mL cup of coffee is too hot to drink when served at 90.0 ∘C.

A)What is the mass of an ice cube, taken from a -16.0 ∘C freezer, that will cool your coffee to a pleasant 58.0 ∘?

Answer 1

The hot water (ignore that it is coffee) will drop from 90 to 58. That's 23 degrees. We will assume all the energy lost will go towards the ice. That calculation for the amount of energy is:

Assume the density of the coffee was 1.0 g/mL. , 300 mL = in 300 g

q = (300 g) (32 C) (4.184 J/gC)

The ice will do three things:

1) heat up from -16 to zero C (2.06 is the specific heat for ice)
2) melt at zero C (334.16 is the enthalpy of vaporization. I picked J/g so I would not have had a conversion to moles of water if I used 6.02 kJ/mol)
3) heat up from zero C to 58 C

Here's the amount of heat required:

[(x)(16)(2.06)] + [(334.16 J/g) (x)] + [(x) (58) (4.184)]

The first square bracket set is the heating of the ice. The second is the melting and the third is the heating up of the liquid water (that used to be ice) to 58 C.

Here's the final set-up:

[(x)(16)(2.06)] + [(334.16 J/g) (x)] + [(x) (58) (4.184)] = (300 g) (32 C) (4.184 J/gC)

x= 65.87 g