"Solving for Temperature - 23°C to -11°C

[ProBasket]

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?

 

[Curious3141]

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?

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.

 

[ProBasket]

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 I am asking for help. i just double checked agian, and got the same answer as i did.

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

 

[Curious3141]

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 I am asking for help. i just double checked agian, and got the same answer as i did.

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

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).

 

[ProBasket]

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 don't get what your trying to tell me

 

[xanthym]

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?

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


~~

 

[ProBasket]

ohhhh... 3 MORE mins... i don't know what was wrong with me...thanks for the help!

 

[Curious3141]

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 don't get what your trying to tell me

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.

And the equations in your latest post are correct. I was wondering about this equation in your 1st post :

23 - (-11) = 34e^{kt}

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} = -1.06deg C

See ? No need to use logs at all.

 

[ProBasket]

the equation from my first post are the same as the equations from my last post. if i subbed in 4 for the equation in my first post, i get the correct answer. i just typed it out differently i guess.

 

[Curious3141]

the equation from my first post are the same as the equations from my last post. if i subbed in 4 for the equation in my first post, i get the correct answer. i just typed it out differently i guess.

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.