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

Answer:

the final temperature = 74.33°C

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 = $\frac{mL_v}{MC}$

where;

m = 2.50 g

M = (240 - 2.50) g  = 237.5 g

$L_v$  = 539 kcal/kg

c = 1.00kcal/kg. °C

ΔT = $\frac{2.50*539 \ kcal /kg}{237.5 g *1.00 \ kcal/kg . ^0C}$

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

Answer 2

Answer:

Final temperature of the hot coffee, $\theta_{f} = 85.67^0 C$

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)

$1347.5 = 237.5 \triangle \theta\\\triangle \theta = 1347.5/237.5\\\triangle \theta =5.67^{0} C$

$\triangle \theta = \theta_{f} - \theta_{i} \\5.67 = \theta_{f} - 80\\\theta_{f} = 80 + 5.67\\\theta_{f} = 85.67^0 C$