The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 250 g of hot coffee initially at 90.0ºC if 2.00 g evaporates from it? The coffee is in a Styrofoam cup, so other methods of heat transfer can be neglected. (answer in ºC)

For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state

Q₁ = m L

Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water

Q₁ = 2.00 10⁻³ 2.26 10⁶

Q1 = 4.52 10³ J

Now the heat of coffee in the cup, which does not change state is

Q coffee = M [tex]c_{e}[/tex] ( [tex]T_{f}[/tex] -[tex]T_{i}[/tex])

Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat

Qc = - Q₁

M ce ([tex]T_{f}[/tex] -[tex]T_{i}[/tex]) = - Q₁

The coffee dough left in the cup after evaporation is

M = 250 -2 = 248 g = 0.248 kg

[tex]T_{f}[/tex] -Ti = -Q1 / M [tex]c_{e}[/tex]

[tex]T_{f}[/tex] = Ti - Q1 / M [tex]c_{e}[/tex]

Since coffee is essentially water, let's use the specific heat of water,

[tex]c_{e}[/tex]= 4186 J / kg ºC

Let's calculate

[tex]T_{f}[/tex] = 90.0 - 4.52 103 / (0.248 4.186 103)

[tex]T_{f}[/tex] = 90- 4.35

[tex]T_{f}[/tex] = 85.65 ° C

[tex]T_{f}[/tex] = 85.7 ° C