• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Further Growth and Decay (1 Viewer)

Kutay

University
Joined
Oct 20, 2004
Messages
599
Location
Castle Hill
Gender
Male
HSC
2005
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?
 

damo676767

Member
Joined
Oct 20, 2004
Messages
149
Location
Winmalee, Blue Mountains
Gender
Male
HSC
2005
the rate that somthings temp chages is proportional tp the out side temp
there fore
T = temp of iron
t = time
A = temp of freazer
K, C are constants

dT/dt = k( (T - A)

by intergratioon we can work out

T = A + Cekt

at t = 0 T = 150
150 = -10 + C
C = 160

T = -10 + 160ekt

at t = 30 T=60
60 = -10 + 160e30k
ek30 = 7/16
t = loge(7/16)/30

T = -10 + 160etloge(7/16)/30

the t = 60

T = -10 + 160e60loge(7/16)/30
= 20.625

at T = 60
0 = -10 + 160etloge(7/16)/30
t = 101 sec approx
 

Bellow

Member
Joined
Oct 3, 2005
Messages
36
Gender
Male
HSC
2005
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
 
Last edited:

Bellow

Member
Joined
Oct 3, 2005
Messages
36
Gender
Male
HSC
2005
damo676767 said:
the rate that somthings temp chages is proportional tp the out side temp
there fore
T = temp of iron
t = time
A = temp of freazer
K, C are constants

dT/dt = k( (T - A)

by intergratioon we can work out

T = A + Cekt

at t = 0 T = 150
150 = -10 + C
C = 160

T = -10 + 160ekt

at t = 30 T=60
60 = -10 + 160e30k
ek30 = 7/16
t = loge(7/16)/30

T = -10 + 160etloge(7/16)/30

the t = 60

T = -10 + 160e60loge(7/16)/30
= 20.625

at T = 60
0 = -10 + 160etloge(7/16)/30
t = 101 sec approx
different answer? ill recheck.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top