What is the constant temperature (M) of the kitchen?

[rad0786]

I need some help.

A pot of boiling water at 100C is removed from a stove at time t = o and left to cool in the kitchen. After 5 min, the water temperature has decreased to 80C, and another 5min later it has dropped to 65C. Assume Newtons law of cooling (dT/dt = k(M - T) ) applies, determain the constant temperature (M) of the kitchen.

Okay, so the solution to the differential equation is - T= Ce^(kt) + M and we want to solve for M

T(0) = C +M
100 = C + M

Equation 1.
T(5) = Ce^(k5) + M
80 = Ce^(k5) + M

Equation 2.
T(10) = Ce^(k10) + M
65 = Ce^(k10) + M

Therefore.. the 2 equations are
80 = Ce^(k5) + M
65 = Ce^(k10) + M

and if I subtract them, I get

15 = Ce^(k5) - Ce^(k10)

And this is where I am lost. I have no idea how to solve for M

Can somebody help me please?

Thanks

 

[saltydog]

So you have:

$$\frac{dT}{dt}=k(M-T)$$

or:

$$\frac{dT}{M-T}=kdt$$

Integrating:

$$-ln(M-T)=kt+C$$

or:

$$ln(M-T)=C-kt$$

(minus c, plus c same dif it's arbitrary)

so that:

$$M-T=e^{C-kt}$$

or:

$$T(t)=M-ce^{-kt}$$

So you have 3 conditions, and 3 unknowns, kinda' messy but try and make those substitutions and then solve for something like M in terms of c, c in terms of k, then substitute that into one of them to solve for k, then the others.

 

[HallsofIvy]

You are given the temperature at 3 times because you have 3 unknown numbers (C, k, and M) and need to solve 3 equations. You wrote the third equation but didn't number it or use it: C+ M= 100.

You can use that to write the two equations 80= Ce^5k+ M and 65= Ce^10k+ M as 80= (100-M)e^5k+ M and 65= (100-M)e^10k+ M. Now, you want to eliminate k, not M, from those equations. Since e^10k= (e^5k)² one way to do that is to write (80- M)²= (100-M)²e^10k and 65- M= (100-M)e^10k and divide one equation by the other. You get
$$\frac{(80-M)^2}{65-M}= 100- M$$.
That's a quadratic equation for M.

 

[rad0786]

Salty dog, when i originally did solved the differential equation, i got the same answer as you, T= M - Ce^(-kt)

However, my textbook solution gives T= Ce^(kt) + M.

Now I am really confused. which one is right?

Well... I did land at the same answer for M for both answers, so i assume they are both correct

 

[Integral]

Since C, k, and M are arbitrary constants their algebraic sign is arbitrary until you have determined their value.

 

[saltydog]

Ok guys. Didn't notice that.

They are the same equation when solved for the constants:

$$T=M-Ce^{-kt};\;M=20,\;C=-80,\;k=0.0575$$

$$T=M+Ce^{kt};\;M=20,\;C=80,\;k=-0.0575$$

 

[rad0786]

okay i have one more differential equation...

m(dv/dt) = mg - bv

where m is the mass, g is the acceleration of gravity, and b>0 is a constant.
if m = 100kg, g = 9.8m/s^2 and b=5 kg/sec, and v(0) = 10m/s, solve for v(t). what is the terminal velocity of the object.

So firstly, i want to solve for this differential equation.
But i cannot find a way to separate it. (i mean, its not seperable?)

m(dv/dt) = mg - bv
(dv/dt) = g - bv/m

then i pluged in the numbers and got

(dv/dt) = 9.8 - 0.05v

and that's about as far as i could get.

could somebody please give me a little push

thanks

 

[qbert]

It is seperable, I'm not sure that's the
way i would solve it but:

(dv/dt) = g - (b/m)v => dv/(g - vb/m)= dt

then a change of variables v -> vb/m turns this
into a simple integral.

 

[rad0786]

oh actually you are right...

dv/(g - vb/m)= dt and you sub in the constants to get

dv/(9.8-0.05v) = dt and solve

(-1/0.05)ln|9.8 - 0.05v| = t + c

and isolate for v to get

v(t) = 196 - 20e^(-0.05t)(c)

and given that v(0) = 10, you can solve for C to get

v(t) = 196 - 186e^(-0.05t)

Now.. the problem asks "What is the limiting (i.e, terminal) velocity of the object?"

Would that just be the limit t-> infinity ? that would get 10m/s?

 

[qbert]

the idea is right:

v_term = limit (t->infinity) 196 - 186 exp(- t/20) = 196.

given v(0) = v0 and m dv/dt = mg - bv we get:
v(t) = gm/b + (v0 - gm/b)exp(-bt/m)

and v_terminal = limit (t-> infinity) v(t).
thus v_terminal = gm/b

[ as lim t->infinity exp(-k*t) = 0 if k>0 ]

we can check the answer, since at v_terminal
there is no acceleration. that is dv/dt = 0 at v_terminal
and the equation of motion is:

m dv/dt = 0 = m g - b v_terminal
=> v_terminal = mg/b.

i always like problems that have built in checks
to see whether or not i get the right answer.