Respuesta :
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[tex]L_v[/tex]
where;
[tex]L_v[/tex] is the latent heat of vaporization.
From the question ;
Let M represent the mass of the coffee that remains after evaporation is:
ΔT = [tex]\frac{mL_v}{MC}[/tex]
where;
m = 2.50 g
M = (240 - 2.50) g = 237.5 g
[tex]L_v[/tex] = 539 kcal/kg
c = 1.00kcal/kg. °C
ΔT = [tex]\frac{2.50*539 \ kcal /kg}{237.5 g *1.00 \ kcal/kg . ^0C}[/tex]
ΔT = 5.67°C
The final temperature of the coffee is:
[tex]T_f = T_i -[/tex] ΔT
where ;
[tex]T_I[/tex] = initial temperature = 80 °C
[tex]T_f[/tex] = (80 - 5.67)°C
[tex]T_f[/tex] = 74.33°C
Thus; the final temperature = 74.33°C
Answer:
Final temperature of the hot coffee, [tex]\theta_{f} = 85.67^0 C[/tex]
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, [tex]\theta_{i} = 80^{0} C[/tex]
Initial temperature of hot coffee, [tex]\theta_{f} =[/tex] ?
Let the specific heat capacity of the coffee, c = 1 kcal/kg
Latent heat of vaporization of coffee, [tex]L_{v} = 539 kcal/kg[/tex]
The heat energy due to temperature change:
[tex]Q = Mc \triangle \theta[/tex]
[tex]Q = 237.5 * 1 * \triangle \theta[/tex]...........(1)
The heat energy due to change in state
[tex]Q = mL_{v}[/tex]
Q = 2.5 * 539
Q = 1347.5..........(2)
Equating (1) and (2)
[tex]1347.5 = 237.5 \triangle \theta\\\triangle \theta = 1347.5/237.5\\\triangle \theta =5.67^{0} C[/tex]
[tex]\triangle \theta = \theta_{f} - \theta_{i} \\5.67 = \theta_{f} - 80\\\theta_{f} = 80 + 5.67\\\theta_{f} = 85.67^0 C[/tex]
