Question

g 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 (in °C) of 240 g of hot coffee initially at 80.0°C if 2.50 g evaporates from it? The coffee is in a Styrofoam cup, and so other methods of heat transfer can be neglected. Assume that coffee has the same physical properties as water.

Answer 1

Final temperature of the hot coffee,
`\theta_(f) = 85.67^0 C`

Step-by-step explanation:

Mass of coffee remaining after evaporation of 2.50 g, M = 240 - 2.50

M = 237.5 g

Mass of evaporated coffee, m = 2.5 g

Initial temperature of hot coffee,
`\theta_(i) = 80^(0) C`

Initial temperature of hot coffee,
`\theta_(f) =` ?

Let the specific heat capacity of the coffee, c = 1 kcal/kg

Latent heat of vaporization of coffee,
`L_(v) = 539 kcal/kg`

The heat energy due to temperature change:

`Q = Mc \triangle \theta`

`Q = 237.5 * 1 * \triangle \theta`...........(1)

The heat energy due to change in state

`Q = mL_(v)`

Q = 2.5 * 539

Q = 1347.5..........(2)

Equating (1) and (2)

`1347.5 = 237.5 \triangle \theta\\\triangle \theta = 1347.5/237.5\\\triangle \theta =5.67^(0) C`

`\triangle \theta = \theta_(f) - \theta_(i) \\5.67 = \theta_(f) - 80\\\theta_(f) = 80 + 5.67\\\theta_(f) = 85.67^0 C`

Answer 2

the final temperature = 74.33°C

Step-by-step explanation:

Using the expression Q = mcΔT for the heat transfer and the change in temperature .

Here ;

Q = heat transfer

m = mass of substance

c = specific heat

ΔT = the change in temperature

The heat Q required to change the phase of a sample mass m is:

Q = m
`L_v`

where;

`L_v` is the latent heat of vaporization.

From the question ;

Let M represent the mass of the coffee that remains after evaporation is:

ΔT =
`(mL_v)/(MC)`

where;

m = 2.50 g

M = (240 - 2.50) g = 237.5 g

`L_v` = 539 kcal/kg

c = 1.00kcal/kg. °C

ΔT =
`(2.50*539 \ kcal /kg)/(237.5 g *1.00 \ kcal/kg . ^0C)`

ΔT = 5.67°C

The final temperature of the coffee is:

`T_f = T_i -` ΔT

where ;

`T_I` = initial temperature = 80 °C

`T_f` = (80 - 5.67)°C

`T_f` = 74.33°C

Thus; the final temperature = 74.33°C