Kutay said:
Hey i was wondering if anyone could help me with understanding this? maybe be showing my how to do this question..
"A hot piece of iron with a temperature of 150degress is placed in a freezer maintained at -10degrees. after 30 seconds the temperature of the iron is 60degrees.
(a) find the temp after 1 minute
(b) when will the temp be 0Degrees?
You know that this is a temperature question so that it relates to Newton's law of cooling which states that the cooling rate of a body is proportional to the difference between the temperature of a body and that of surrounding medium.
i.e dT/dt = -k(T-M)
where T is the temperature at any time t and M is the temperature of the surrounding medium.
so from the question u kno wat M is....the freezer which is fixed at -10 degrees
.: M = -10
dT/dt = -k(T + 10)
reciprocate both sides and make dt the subject
dt = -dT/[k(T+10)]
integrate both sides
∫dt = -1/k x ∫1/(T+10) dT (note that k is a constant)
t = -1/k x ln(T+10) + C
rearrange the equation so that T is the subject
-k(t-C) = ln(T+10)
-kt + kc = ln(T+10)
e^(-kt + kC) = ln(T+10)
e^-kt x e^kc = T+10
T = e^-kc x e^kc - 10 now let e^kc be any constant, say A
.: T = -10 + Ae^-kt
(i)T = -10 + Ae^-kt
when t = 0, T = 150 (as the hot plate is originally at 150 degrees)
150 = -10 + Ae^-k(0)
150 = -10 + A(1)
A = 160
T = -10 + 160e^-kt
when t = 30seconds, T = 60 degrees
60 = -10 + 160e^-k(30)
rearrange to find k
70/160 = e^-30k
ln[70/160] = -30k
k = 0.02755......
so afta 1 minute, i.e t=60seconds
T = -10 +160e^-0.02755(60)
.: T = 21 degrees
therefore the temperature afta one minute is 21 degrees
(ii) T = -10 + 160e^-0.02755t
When T = 0 (when the temp. drops to 0)
0 = -10 + 160e^-0.02755t
10/160 = e^-0.02755t
ln[1/16] = -0.02755t
.: t = 101 seconds
therefore the time it takes for the temperature to be 0 is 101 seconds
Edit: sorry made a couple of errors
