Respuesta :
Explanation:
The given data is as follows.
Initial temperature ([tex]T_{1}[/tex]) = [tex]50^{o}C[/tex]
Room temperature ([tex]T_{a}[/tex]) = [tex]20^{o}C[/tex]
Time (t) = 5 min
Final temperature (T) = [tex]40^{o}C[/tex]
According to the unsteady state equation,
[tex]\frac{ln (T - T_{a})}{(T_{1} - T_{a})} = \frac{-h \times A \times t}{dV \times C}[/tex]
where, h = heat transfer coefficient
A = area
d = density
V = volume
C = specific heat capacity
All of these are constants and can be expressed as K
Therefore, K = [tex]\frac{h \times A}{dV \times C}[/tex]
[tex]\frac{ln (T - T_{a})}{(T_{1} - T_{a})} = -K \times t[/tex]
[tex]\frac{ln (40 - 20)}{(50 - 20)} = -K \times 5[/tex] 0.4055 = 5K
K = 0.0811
After 5 minutes, the new temperature will be as follows.
New temperature ([tex]T_{2}[/tex]) = [tex]30^{o}C[/tex]
Time (t) = ?
Again, from the unsteady state conduction,
[tex]\frac{ln (T_{2} - T_{a})}{(T_{i} - T_{a})} = [tex]\frac{-h \times A \times t}{dV \times C}[/tex] = [tex]-K \times t[/tex]
[tex]\frac{ln [(30 - 20)}{(50 - 20)} = - 0.0811 \times t[/tex]
1.0986 = [tex]0.0811 \times t[/tex]
t = 13.55 min
Thus, we can conclude that after 13.55 minutes the temperature reaches to [tex]30^{o}C[/tex].
