Now this is a first order ordinary differential

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Now this is a first order ordinary differential equation for T ( t ). We can separate variables to get dT T T = h A ρcV dt, (5.188) integraldisplay dT T T = integraldisplay h A ρcV dt, (5.189) ln( T T ) = h A ρcV t + C, (5.190) T T = C exp parenleftbigg h A ρcV t parenrightbigg . (5.191) CC BY-NC-ND. 2011, J. M. Powers.
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5.6. TIME-DEPENDENCY 133 Here C = exp( C ) and is also a constant. Now when t = 0, we have T = T o , so T o T = C exp parenleftbigg h A ρcV 0 parenrightbigg , (5.192) = C . (5.193) Thus T ( t ) = T + ( T o T ) exp parenleftbigg h A ρcV t parenrightbigg . (5.194) Notice that T (0) = T o and T ( ) = T . And notice that when the argument of the exponential is 1, giving exp( 1) = 0 . 357, we get a good estimate of the time it takes to get most of the way to equilibrium. We define this as the time constant , τ of the system. Here, we can get τ via h A ρcV τ = 1 , (5.195) τ = ρcV h A . (5.196) So we get fast cooking (small τ ) if ρ is small (a light potato) c is small, h is large (the heat transfer rate is fast), A/V is large (the surface to volume ratio is large). For our potato, let us model it as a sphere of liquid water with R = 0 . 05 m , ρ = 997 kg/m 3 , c = 4 . 18 kJ/kg/K . Let us take h = 0 . 012 kW/m 2 /K . Let us take T o = 20 C , T = 200 C . For a sphere, we have the surface to volume ratio of A V = 4 π R 2 (4 / 3) π R 3 = 3 R . (5.197) Our temperature varies with time as T ( t ) = T + ( T o T ) exp parenleftbigg 3 h ρc R t parenrightbigg , (5.198) = (20 C ) + ((20 C ) (200 C )) exp 3 ( 0 . 012 kW m 2 K ) parenleftBig 997 kg m 3 parenrightBigparenleftBig 4 . 18 kJ kg K parenrightBig (0 . 05 m ) t , (5.199) T ( t ) = (200 C ) (180 C ) exp parenleftbigg t 5788 . 14 s parenrightbigg (5.200) The time constant τ = 5788 . 14 s = 96 . 469 min . When T = 100 C , the potato is probably cooked enough. This occurs at t = 3402 . 19 s parenleftbigg min 60 s parenrightbigg = 56 . 7032 min. (5.201) If we leave the potato in the oven too long, it will get too hot, and all its water will boil away. The temperature history is plotted in Fig. 5.18. This example used what is known as the lumped capacitance method for analysis of T ( t ). In actuality, T is a function of space and time: T = T ( x, y, z, t ). It will be shown in later courses that the lumped capacitance method is valid when the so-called Biot number , Bi , is much less than unity. Biot himself CC BY-NC-ND. 2011, J. M. Powers.
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134 CHAPTER 5. THE FIRST LAW OF THERMODYNAMICS 0 5000 10000 15000 20000 25000 30000 50 100 150 200 T (˚C) t (s) t = 3402.19 s = 56.7 min cooking time Figure 5.18: Temperature versus time for the potato in the oven. was an important physicist who played a small role in thermal sciences. He is depicted in Fig. 5.19. The Biot number for our potato problem is defined as Bi hR k , (5.202) where k is the thermal conductivity. For the lumped capacitance to be valid, we need Bi = hR k << 1 , thus k >> hR . (5.203) For liquid water, k 0 . 006 kW/m/K , so for this problem Bi = ( 0 . 012 kW m 2 k ) (0 . 05 m ) 0 . 006 kW m K = 0 . 1 . (5.204) This is a small enough Biot number that our lumped capacitance method is acceptable. Physically, if the thermal conductivity is high relative to the product of h and R , thermal energy diffuses rapidly in the solid, and internal temperature gradients are small.
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