Newtons laws of cooling proposes that the rate of change of temperature is proportional to the temperature difference to the ambient (room) temperature. And can be modelled using the equation: dT/dt = -k (T-Ta)
It can also be written as dT/T-Ta = -k dt
Where:
T = Temperature of material
Ta = Ambient (room) temperature
k = A cooling constant
a) integrate both sides of the equation and show that the temperature difference is given by:
(T-Ta) = CoE^-kt
(Co is a constant for this problem)
B) calculate Co if the initial temperature is 70 degrees C and Ta = 20 degrees C?
Dear Teresa Judge-Bird, in the integral of k=k*t ^0 the formulae
∫kf(t)dt=k ∫f(t)dt and ∫t^0dt=t^(0+1)/(0+1)+C1 were applied. The
integration constant C1 was denoted by ln C_0 in a solution of the
question.
Teresa Judge-Bird
26.04.21, 03:43
Hi I am struggling to understand how you went from ∫kdt to kt+lnC0?
Assignment Expert
14.01.21, 23:24
The variables T and t were separated, the corresponding left-hand and
right-hand sides of a new differential equation were integrated. Next,
the logarithmic equation was solved. The absolute value sign can be
omitted using the sign of C_0 in a solution of the question.
Khalid Makki
14.01.21, 23:15
can you explain more how did you answer part a?
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Dear Teresa Judge-Bird, in the integral of k=k*t ^0 the formulae ∫kf(t)dt=k ∫f(t)dt and ∫t^0dt=t^(0+1)/(0+1)+C1 were applied. The integration constant C1 was denoted by ln C_0 in a solution of the question.
Hi I am struggling to understand how you went from ∫kdt to kt+lnC0?
The variables T and t were separated, the corresponding left-hand and right-hand sides of a new differential equation were integrated. Next, the logarithmic equation was solved. The absolute value sign can be omitted using the sign of C_0 in a solution of the question.
can you explain more how did you answer part a?
Leave a comment