In: Math
A pizza pan is removed at 4:00 pm from an oven whose
temperature is fixed at 450 F into a room that is constant 73 F.
after 5 minutes, the pizza is at 300 F.
a)at what time is the temperature of the pan 125 F?
b) determine the time that needs to elapse before the pans is 180
degree .
c)what do you notice about the temperature as time passes?
The temperature difference when a hot object is placed into a colder environment is a decreasing exponential function of the form D(t) = c*e-kt, where c is the initial temperature difference (at t = 0) , t is the time measured in minutes, D(t) is the temperature of the hot object after t minutes and k > 0 is a constant of heat exchange. Here, c = 450-73 = 377. Also, 300 = 377e-5k. On taking natural logarithm of both the sides, we have ln 300 = ln(377e-5k) = ln 377 -5k ln e = ln 377 -5k so that 5k = ln 377-ln 300 = 5.932245187-5.703782475 = 0.228462712. Hence k = 0.228462712/5 = 0.045692542.
(a).Let the temperature of the pan be 125o F, t minutes after 4.00 p.m. Then 125 = 377e-0.045692542t. On taking natural logarithm of both the sides, we have ln 125 = ln 377-0.045692542t or, t = (ln 377-ln125)/ 0.045692542 = (5.932245187-4.828313737)/ 0.045692542 = 1.10393145/0.045692542 = 24.15999201 minutes or, 24.16 minutes ( on rounding off to 2 decimal places).
(b).Let the temperature of the pan be 180o F, t minutes after 4.00 p.m. Then 180 = 377e-0.045692542t. On taking natural logarithm of both the sides, we have ln 125 = ln 377-0.045692542t or, t = (ln 377-ln180)/ 0.045692542 = (5.932245187- 5.192956851)/ 0.045692542 = 0.739288336/0.045692542 = 16.18 minutes ( on rounding off to 2 decimal places).
(c).The temperature of the pizza pan declines with the passage of time.