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

Answer:

$T_{f}$ = 85.7 ° C

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