# Homework Help: Temperatures - math

1. Mar 19, 2005

A thermometer is taken from a room where the temperature is $$23^{o}C$$ to the outdoors, where the temperature is $$-11^{o}C$$. After one minute the thermometer reads $$14^{o}C$$.

A.) What will the reading on the thermometer be after 3 more minutes?

my work:
$$T=23^{o}C$$
$$T_s = -11^{o}C$$
$$23 - (-11) = 34e^{kt}$$ <-- initial temp right?
after 1 min...
$$t(1) = 34e^{k1} = 14 - (-11)$$
$$t(1) = 34e^{k1} = 25$$
solved for k and got -0.307484
my equation:
$$y(0) = 34e^{-0.307484*t} - 11$$

so i just subbed in 3 for t and solved and got 2.516, but it's incorrect. anyone know where i went wrong?

2. Mar 19, 2005

### Curious3141

Hint 1 :

The equation for Newtonian cooling is $$T(t) = T_s + (T_0 - T_s)e^{-kt}$$

where $$T(t)$$ is the temperature of the thermometer at time t, $$T_s$$ is the temp of the surroundings, $$T_0$$ is initial temp of the thermometer and $$t$$ is time.

Your equation for the "initial conditions" makes no sense.

Hint 2 : 3 more minutes.

3. Mar 19, 2005

i did follow that rule, didnt i? well that was the equation that i was using to get my intial condition, unless i went wrong somewhere, which is where im asking for help. i just double checked agian, and got the same answer as i did.

T(0) = 23
T_s = -11
right?

4. Mar 19, 2005

### Curious3141

You used the equation wrongly. The values you are substituting for initial temp and surrounding temp are correct.

Do you need to set up one equation for the initial temp ?

Set up the correct equation for t = 1 minute. What is T(1) ?

What is the t-value at 3 more minutes ? (the italics are a big hint).

5. Mar 19, 2005

T(1) = 14

$$T(t) = T_s + (T_0 - T_s)e^{-kt}$$

$$14 = -11 + (23- (-11))e^{-kt}$$
$$25 = 34e^{-kt}$$ which is what i got

for 3 min later, i would just sub in 3 for t right? i really dont get what your trying to tell me

6. Mar 19, 2005

### xanthym

Mathematical model:
(dT/dt) = k*(T - Ts)
where Ts is the fixed outside temp. Solving and using Ts=(-11 degC), we get:
T(t) = A*exp(k*t) + Ts =
= A*exp(k*t) - 11

From the problem statement:
T(0) = 23 = A*exp{k*(0)} - 11 =
= A - 11
::: ⇒ A = (34)
::: ⇒ T(t) = (34)*exp(k*t) - 11

T(1) = 14 = (34)*exp{k*(1)} - 11
::: ⇒ exp{k} = (14 + 11)/34 = (0.735294)
::: ⇒ k = Loge(0.735294) = (-0.307485)
::: ⇒ T(t) = (34)*exp{(-0.307485)*t} - 11

T(1 + 3) = T(4) = (34)*exp{(-0.307485)*(4)} - 11
T(4) = (-1.0615 degC)

~~

Last edited: Mar 19, 2005
7. Mar 19, 2005

ohhhh.... 3 MORE mins..... i dont know what was wrong with me...thanks for the help!!!

8. Mar 19, 2005

### Curious3141

Well, xanthym has solved the question completely, so I guess it's moot now.

I was saying for "3 minutes later" you should use t = 4. You see why right ? Must be careful in reading the question.

which I can't figure out how you got.

At any rate, a much "neater" way of solving the question without computing k directly is to find the value of $$e^{-k}$$ like so :

$$25 = 34e^{-k}$$

$$e^{-k} = \frac{25}{34}$$

At t = 4,

$$T(4) = -11 + 34e^{-4k}$$

$$T(4) = -11 + 34{(e^{-k})}^4$$

$$T(4) = -11 + (34){(\frac{25}{34})}^4 = -11 + \frac{25^4}{34^3} = -1.06$$deg C

See ? No need to use logs at all.

9. Mar 19, 2005

Ohh, I see. You were just trying to say $$34 = 34e^{0}$$. OK, get it now. I just thought that bit was obvious. Never mind, then you have the correct method.