A turkey has an initial temperature of 40 degrees Celsius. It is placed in a 450 degree oven, and after one hour its temperature is 90 degrees Celsius. What is the turkey's temperature as a function of time?
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Let T(t) be the temperature as a function of time t > = 0 in hours
We have T(0) = 40, T_a = 450 (ambient temperature) and T(1) = 90.
According to Newton's Law of Cooling: dT(t) / dt = - k (T - T_a)
Solving this diff. equation we get T(t) = T_a + (T(0) - T_a) exp( - kt)
Let t = 1 and use T(1) = 90, T_a = 450 to find k. Then use k to find T(t).
We have T(0) = 40, T_a = 450 (ambient temperature) and T(1) = 90.
According to Newton's Law of Cooling: dT(t) / dt = - k (T - T_a)
Solving this diff. equation we get T(t) = T_a + (T(0) - T_a) exp( - kt)
Let t = 1 and use T(1) = 90, T_a = 450 to find k. Then use k to find T(t).
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decay function : A=A0*e^(-kt)
growth function: A=A0*e^(kt)
in this case we have decay so we can write
T=T0*e^(-kt)
where
T= temperature after time t
T0=initial temperature (450deg)
k is some constant
the problem is that we don't know k, so we can solve for k and use known data:
T/T0=e^-kt
ln(T/T0)=-kt
k=ln(T0/T) / t
k=ln(450/90)/1
k=ln(5) ...... (= 1.6094379124341003746007593332262)
back to equation
T=T0*e^(-ln(5) * t)
T=T0*e^( ln(0.2) * t)
T=450*e^( ln(0.2) * t)
where t is time in hours (because k is calculated for t=1h, for other time units, recalculate k)
growth function: A=A0*e^(kt)
in this case we have decay so we can write
T=T0*e^(-kt)
where
T= temperature after time t
T0=initial temperature (450deg)
k is some constant
the problem is that we don't know k, so we can solve for k and use known data:
T/T0=e^-kt
ln(T/T0)=-kt
k=ln(T0/T) / t
k=ln(450/90)/1
k=ln(5) ...... (= 1.6094379124341003746007593332262)
back to equation
T=T0*e^(-ln(5) * t)
T=T0*e^( ln(0.2) * t)
T=450*e^( ln(0.2) * t)
where t is time in hours (because k is calculated for t=1h, for other time units, recalculate k)