Question

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)

Answer 1

`T_(f)` = 85.7 ° C

Step-by-step explanation:

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
`c_(e)` (
`T_(f)` -
`T_(i)`)

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

Qc = - Q₁

M ce (
`T_(f)` -
`T_(i)`) = - Q₁

The coffee dough left in the cup after evaporation is

M = 250 -2 = 248 g = 0.248 kg

`T_(f)` -Ti = -Q1 / M
`c_(e)`

`T_(f)` = Ti - Q1 / M
`c_(e)`

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

`c_(e)`= 4186 J / kg ºC

Let's calculate

`T_(f)` = 90.0 - 4.52 103 / (0.248 4.186 103)

`T_(f)` = 90- 4.35

`T_(f)` = 85.65 ° C

`T_(f)` = 85.7 ° C