Respuesta :
Answer: t = 9.05 min
Step-by-step explanation: The rate of temperature over time is given by:
[tex]\frac{dT}{dt} = k (T - T_c)[/tex]
where:
K is the constant of transference of heat
[tex]T_c[/tex] is temperature of the container
[tex]\int\limits^a_b {\frac{1}{T-T_c} } \, dT = \int\limits^a_b {k} \, dt[/tex]
ln (T - [tex]T_c[/tex]) = kt + c
[tex]e^{kt+c} = T - T_c[/tex]
[tex]Ce^{kt} = T - T_c[/tex]
[tex]T = Ce^{kt} + T_c[/tex]
In container A, the temperature is at 0°C, so
[tex]T = Ce^{kt} + 0[/tex]
For the bar, when t = 0 min, T = 100°C:
[tex]T = Ce^{kt}[/tex]
100 = [tex]Ce^{k.0}[/tex]
C = 100
After 1 minute, the temperature of the bar is 90°C, so:
[tex]T = 100e^{kt}[/tex]
[tex]100e^{k.1} = 90[/tex]
[tex]e^{k} = \frac{90}{100}[/tex]
[tex]ln (e^k) = ln (0.9)[/tex]
k = ln(0.9)
k = - 0.105
[tex]T = 100e^{-0.105t}[/tex]
After 2 minutes, the temperature will be:
[tex]T = 100e^{-0.105.2}[/tex]
T = 81.06°C
For container B, the temperature is 100°C, so:
[tex]T = Ce^{kt} + T_c[/tex]
[tex]T = Ce^{kt} + 100[/tex]
The initial temperature of the bar when entering the container B is T = 81.06°C, then:
[tex]81.06 = Ce^{k.0} + 100[/tex]
C = - 18.94
After 1 minute, the temperature rises 10°C:
[tex]T = - 18.94e^{kt} + 100[/tex]
91 = [tex]- 18.94e^{k.1} + 100[/tex]
[tex]e^{k} = \frac{9}{18.94}[/tex]
[tex]e^{k} = 0.475[/tex]
k = ln(0.475)
k = - 0.744
[tex]T = - 18.94e^{-0.744t} + 100[/tex]
When T = 99.9°C:
[tex]99.9 = - 18.94e^{-0.744t} + 100[/tex]
[tex]e^{- 0.744t} = \frac{99.9 - 100}{- 18.94}[/tex]
[tex]e^{-0.744t} = 0.0053[/tex]
t = [tex]\frac{ln(0.0053)}{-0.744}[/tex]
t = 7.05 min
The entire process will take:
t = 2 + 7.05
t = 9.05 min
