Differential equations help?

You've got two unknowns in this problem, the constant k, and that of integration (I'll call it C).

I'm going to let T be tau ro simplify

You've already got the differential equation correct

dT/dt=-k(T-To)

with the following boundary (and other) conditions

e.g. at T=96 deg dT/dt=-1 deg/t and the boundary condition that t=0 T=96 deg and To=4 deg

Use the first condition to get k

-1 deg/t=-k(96-4)deg

rearranging to solve for k

k=(-1deg/t)/(96-4)deg=1/92t

Now the differential equation becomes

dT/dt=-(1/92)(T-4)

separating the variables (dT and dt)

dT/(T-4)=-(1/92)dt

Integrating

ln((T-4)=-(1/92)t+C

where C is a constant of integration

Putting in the boundary conditions (@t=0 T=96)

Ln(96-4)=Ln(92)=[-(1/92)*0+C]
&
C=ln(92)

so Ln(T-4)=-(1/92)t+Ln(92)

puttin in the antilog

T-4=exp(-(1/92)t+ln(92))==exp(Ln(92))*exp(-t/92)=92exp(-t/92)

T=4+92exp(-t/92)

so now I got this simple equation that gives me object temperature (T) with respect to time (t)

but first I wanna check my answer to see if it makes sense, so I'll check the boundary conditions (@t=0 T=96)

T=4+92exp(-0/92)=4+92=96

Bingo

Rap

ah thnx raprap! dats veri helpful tho im confused at the very first part....how did you get -1 deg/t?

Also I have just attempted a similar question which relates to radioactive decay. I have to determine the half-life of radioactive atoms when given that the specific rate of decay is 2x 10^(-5) - which I got = 0.00002
The answer I got for the half life was 34657.35903 seconds, is that about right? The last part of the question ive been asked to determine how long it will be until 1% of the original radioactive atoms remain. Instead of finding half of the atoms, I set it to 0.01 and I got the answer 230258.5093 seconds........

But if its asking me how long UNTIL only 1% remain, does that mean I take the difference of my two answers?

Note: There isn't a given value of original atoms, just have to solve like a general formula

That is from dT/dt which from what you gave me is that the temperature is decreasing (the minus) 1 degree per unit time (t). You didn't say what the unit was (second, minute, hour, year, decade, century)---whatever the time base was dT/dt=-1 (deg/t).

Radioactive decay is similar. If the initial mass is Mo the amount you have is

M=Mo[exp(-lambda*t)] where lambda is the specific rate of decay.

The half life is the amount of time (t1/2) required for half of the material to decay.

So

Mo/2=Mo[exp(-lambda*t1/2)] and you solve for t1/2

or

1/2=exp(-lambda*t1/2)

ln(1/2)=-lambda*t1/2

and

t1/2=ln(1/2)/(-lmbda)=ln2/lmbda

Similarly

lambda=ln(2)/t1/2

The determination of amount of time required for 1% (0.01) of material left is similar to the half life.

0.01Mo=Mo[exp(-lambda*t')]

and solve for t'

Hint (like the half life the Mo's cancel)

Cautions-some isotopes lambda's aren't in seconds--they can range from microseconds to billions of years) and when you calculate lambdas from half life make sure the units of time are the same. And ln is the natural logarithm

Rap

That is from dT/dt which from what you gave me is that the temperature is decreasing (the minus) 1 degree per unit time (t). You didn't say what the unit was (second, minute, hour, year, decade, century)---whatever the time base was dT/dt=-1 (deg/t).

Radioactive decay is similar. If the initial mass is Mo the amount you have is

M=Mo[exp(-lambda*t)] where lambda is the specific rate of decay.

The half life is the amount of time (t1/2) required for half of the material to decay.

So

Mo/2=Mo[exp(-lambda*t1/2)] and you solve for t1/2

or

1/2=exp(-lambda*t1/2)

ln(1/2)=-lambda*t1/2

and

t1/2=ln(1/2)/(-lmbda)=ln2/lmbda

Similarly

lambda=ln(2)/t1/2

The determination of amount of time required for 1% (0.01) of material left is similar to the half life.

0.01Mo=Mo[exp(-lambda*t')]

and solve for t'

Hint (like the half life the Mo's cancel)

Cautions-some isotopes lambda's aren't in seconds--they can range from microseconds to billions of years) and when you calculate lambdas from half life make sure the units of time are the same. And ln is the natural logarithm

Rap

Ah, I get it now! Thanks again Rap