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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)

User Rjss
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1 Answer

4 votes

Answer:


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

User Papershine
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