Mixing
Newton’s Law of Cooling/Warming
$begingroup$
A thermometer is placed in an oven preheated to a constant temperature of 390◦ F.
Through a glass window in the oven door, an observer records that the thermometer
reads 190◦ F after 1 minute and 230◦ F after 2 minutes. What is the initial reading of
the thermometer?
I know that you have to use the formula $frac{dT}{dt}=k(T-T_m)$. I don't know what to do next.
ordinary-differential-equations
asked Mar 24 '15 at 14:29
JessieJessie
283
$endgroup$
add a comment |
$begingroup$
A thermometer is placed in an oven preheated to a constant temperature of 390◦ F.
Through a glass window in the oven door, an observer records that the thermometer
reads 190◦ F after 1 minute and 230◦ F after 2 minutes. What is the initial reading of
the thermometer?
I know that you have to use the formula $frac{dT}{dt}=k(T-T_m)$. I don't know what to do next.
ordinary-differential-equations
asked Mar 24 '15 at 14:29
JessieJessie
283
$endgroup$
$begingroup$
Do you know how to solve the differential equation to get temperature as a function of time? I'm trying to understand which part of this solution you need help with.
$endgroup$
– dshapiro
Mar 24 '15 at 14:33
$begingroup$
One hint is that the value of k remains the same. Equate those values to get T0.
$endgroup$
– Krishna
Mar 24 '15 at 14:39
add a comment |
$begingroup$
A thermometer is placed in an oven preheated to a constant temperature of 390◦ F.
Through a glass window in the oven door, an observer records that the thermometer
reads 190◦ F after 1 minute and 230◦ F after 2 minutes. What is the initial reading of
the thermometer?
I know that you have to use the formula $frac{dT}{dt}=k(T-T_m)$. I don't know what to do next.
ordinary-differential-equations
asked Mar 24 '15 at 14:29
JessieJessie
283
$endgroup$
A thermometer is placed in an oven preheated to a constant temperature of 390◦ F.
Through a glass window in the oven door, an observer records that the thermometer
reads 190◦ F after 1 minute and 230◦ F after 2 minutes. What is the initial reading of
the thermometer?
I know that you have to use the formula $frac{dT}{dt}=k(T-T_m)$. I don't know what to do next.
ordinary-differential-equations
ordinary-differential-equations
asked Mar 24 '15 at 14:29
JessieJessie
283
asked Mar 24 '15 at 14:29
JessieJessie
283
asked Mar 24 '15 at 14:29
JessieJessie
283
asked Mar 24 '15 at 14:29
JessieJessie
283
asked Mar 24 '15 at 14:29
JessieJessie
283
283
$begingroup$
Do you know how to solve the differential equation to get temperature as a function of time? I'm trying to understand which part of this solution you need help with.
$endgroup$
– dshapiro
Mar 24 '15 at 14:33
$begingroup$
One hint is that the value of k remains the same. Equate those values to get T0.
$endgroup$
– Krishna
Mar 24 '15 at 14:39
add a comment |
$begingroup$
Do you know how to solve the differential equation to get temperature as a function of time? I'm trying to understand which part of this solution you need help with.
$endgroup$
– dshapiro
Mar 24 '15 at 14:33
$begingroup$
One hint is that the value of k remains the same. Equate those values to get T0.
$endgroup$
– Krishna
Mar 24 '15 at 14:39
$begingroup$
Do you know how to solve the differential equation to get temperature as a function of time? I'm trying to understand which part of this solution you need help with.
$endgroup$
– dshapiro
Mar 24 '15 at 14:33
$begingroup$
Do you know how to solve the differential equation to get temperature as a function of time? I'm trying to understand which part of this solution you need help with.
$endgroup$
– dshapiro
Mar 24 '15 at 14:33
$begingroup$
One hint is that the value of k remains the same. Equate those values to get T0.
$endgroup$
– Krishna
Mar 24 '15 at 14:39
$begingroup$
One hint is that the value of k remains the same. Equate those values to get T0.
$endgroup$
– Krishna
Mar 24 '15 at 14:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Newton's law of cooling would lead you to the differential equation,
$$frac{dT}{dt} = k(T-T_i)$$
This is a very simple differential equation, which could be solved as:
$$ frac{dT}{T-T_i} = k dt$$
$$ Rightarrow int_0^{t'} frac{dT}{T-T_i} = int_0^{t'} k dt$$
$$ Rightarrow lnleft(frac{T}{T_i}right) = k t'$$
$$ Rightarrow T = T_i exp (k t')$$
Now, the thing with exponential functions is, any point for which you know the values of $(t, T(t))$ can be taken as a starting point. I leave the rest to you as a simple exercise.
answered Mar 24 '15 at 14:45
299792458299792458
378515
$endgroup$
$begingroup$
What does the answer come out to?
$endgroup$
– Jessie
Mar 24 '15 at 16:57
add a comment |
$begingroup$
Newton's law serves equally for cooling or warming situations:
$frac{dT}{dt}=k(T_e-T)$
The general solution is $T(t)=T_e+(T_0-T_e)e^{-kt}$; where $T_0=T(0)$
is the initial reading we are going to find
In this case $T_e=390^{circ}F$ is the oven temperature; so we can write
$T(t)=390+(T_0-390)e^{-kt}$
Now, we have recorded two temperatures at some given moments:
$190=390+(T_0-390)e^{-k(1)}$
$230=390+(T_0-390)e^{-k(2)}$
We have to solve these two last equations to find $T_0$ and $k$
$(T_0-390)e^{-k(1)}=-200$
$(T_0-390)e^{-k(2)}=-160$
$e^{-k}=frac{200}{390-T_0}$
$e^{-2k}=frac{160}{390-T_0}$
$-k=lnfrac{200}{390-T_0}$
$-2k=lnfrac{160}{390-T_0}$
$lnfrac{160}{390-T_0}=2lnfrac{200}{390-T_0}$
$lnfrac{160}{390-T_0}=ln(frac{200}{390-T_0})^2$
$frac{160}{390-T_0}=(frac{200}{390-T_0})^2$
$390-T_0=frac{200^2}{160}=250$
$T_0=140^{circ}F$
With this value we can find $k=ln(1.25)$
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Newton's law of cooling would lead you to the differential equation,
$$frac{dT}{dt} = k(T-T_i)$$
This is a very simple differential equation, which could be solved as:
$$ frac{dT}{T-T_i} = k dt$$
$$ Rightarrow int_0^{t'} frac{dT}{T-T_i} = int_0^{t'} k dt$$
$$ Rightarrow lnleft(frac{T}{T_i}right) = k t'$$
$$ Rightarrow T = T_i exp (k t')$$
Now, the thing with exponential functions is, any point for which you know the values of $(t, T(t))$ can be taken as a starting point. I leave the rest to you as a simple exercise.
answered Mar 24 '15 at 14:45
299792458299792458
378515
$endgroup$
$begingroup$
What does the answer come out to?
$endgroup$
– Jessie
Mar 24 '15 at 16:57
add a comment |
$begingroup$
Newton's law of cooling would lead you to the differential equation,
$$frac{dT}{dt} = k(T-T_i)$$
This is a very simple differential equation, which could be solved as:
$$ frac{dT}{T-T_i} = k dt$$
$$ Rightarrow int_0^{t'} frac{dT}{T-T_i} = int_0^{t'} k dt$$
$$ Rightarrow lnleft(frac{T}{T_i}right) = k t'$$
$$ Rightarrow T = T_i exp (k t')$$
Now, the thing with exponential functions is, any point for which you know the values of $(t, T(t))$ can be taken as a starting point. I leave the rest to you as a simple exercise.
answered Mar 24 '15 at 14:45
299792458299792458
378515
$endgroup$
$begingroup$
What does the answer come out to?
$endgroup$
– Jessie
Mar 24 '15 at 16:57
add a comment |
$begingroup$
Newton's law of cooling would lead you to the differential equation,
$$frac{dT}{dt} = k(T-T_i)$$
This is a very simple differential equation, which could be solved as:
$$ frac{dT}{T-T_i} = k dt$$
$$ Rightarrow int_0^{t'} frac{dT}{T-T_i} = int_0^{t'} k dt$$
$$ Rightarrow lnleft(frac{T}{T_i}right) = k t'$$
$$ Rightarrow T = T_i exp (k t')$$
Now, the thing with exponential functions is, any point for which you know the values of $(t, T(t))$ can be taken as a starting point. I leave the rest to you as a simple exercise.
answered Mar 24 '15 at 14:45
299792458299792458
378515
$endgroup$
Newton's law of cooling would lead you to the differential equation,
$$frac{dT}{dt} = k(T-T_i)$$
This is a very simple differential equation, which could be solved as:
$$ frac{dT}{T-T_i} = k dt$$
$$ Rightarrow int_0^{t'} frac{dT}{T-T_i} = int_0^{t'} k dt$$
$$ Rightarrow lnleft(frac{T}{T_i}right) = k t'$$
$$ Rightarrow T = T_i exp (k t')$$
Now, the thing with exponential functions is, any point for which you know the values of $(t, T(t))$ can be taken as a starting point. I leave the rest to you as a simple exercise.
answered Mar 24 '15 at 14:45
299792458299792458
378515
answered Mar 24 '15 at 14:45
299792458299792458
378515
answered Mar 24 '15 at 14:45
299792458299792458
378515
answered Mar 24 '15 at 14:45
299792458299792458
378515
378515
$begingroup$
What does the answer come out to?
$endgroup$
– Jessie
Mar 24 '15 at 16:57
add a comment |
$begingroup$
What does the answer come out to?
$endgroup$
– Jessie
Mar 24 '15 at 16:57
$begingroup$
What does the answer come out to?
$endgroup$
– Jessie
Mar 24 '15 at 16:57
$begingroup$
What does the answer come out to?
$endgroup$
– Jessie
Mar 24 '15 at 16:57
add a comment |
$begingroup$
Newton's law serves equally for cooling or warming situations:
$frac{dT}{dt}=k(T_e-T)$
The general solution is $T(t)=T_e+(T_0-T_e)e^{-kt}$; where $T_0=T(0)$
is the initial reading we are going to find
In this case $T_e=390^{circ}F$ is the oven temperature; so we can write
$T(t)=390+(T_0-390)e^{-kt}$
Now, we have recorded two temperatures at some given moments:
$190=390+(T_0-390)e^{-k(1)}$
$230=390+(T_0-390)e^{-k(2)}$
We have to solve these two last equations to find $T_0$ and $k$
$(T_0-390)e^{-k(1)}=-200$
$(T_0-390)e^{-k(2)}=-160$
$e^{-k}=frac{200}{390-T_0}$
$e^{-2k}=frac{160}{390-T_0}$
$-k=lnfrac{200}{390-T_0}$
$-2k=lnfrac{160}{390-T_0}$
$lnfrac{160}{390-T_0}=2lnfrac{200}{390-T_0}$
$lnfrac{160}{390-T_0}=ln(frac{200}{390-T_0})^2$
$frac{160}{390-T_0}=(frac{200}{390-T_0})^2$
$390-T_0=frac{200^2}{160}=250$
$T_0=140^{circ}F$
With this value we can find $k=ln(1.25)$
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
$endgroup$
add a comment |
$begingroup$
Newton's law serves equally for cooling or warming situations:
$frac{dT}{dt}=k(T_e-T)$
The general solution is $T(t)=T_e+(T_0-T_e)e^{-kt}$; where $T_0=T(0)$
is the initial reading we are going to find
In this case $T_e=390^{circ}F$ is the oven temperature; so we can write
$T(t)=390+(T_0-390)e^{-kt}$
Now, we have recorded two temperatures at some given moments:
$190=390+(T_0-390)e^{-k(1)}$
$230=390+(T_0-390)e^{-k(2)}$
We have to solve these two last equations to find $T_0$ and $k$
$(T_0-390)e^{-k(1)}=-200$
$(T_0-390)e^{-k(2)}=-160$
$e^{-k}=frac{200}{390-T_0}$
$e^{-2k}=frac{160}{390-T_0}$
$-k=lnfrac{200}{390-T_0}$
$-2k=lnfrac{160}{390-T_0}$
$lnfrac{160}{390-T_0}=2lnfrac{200}{390-T_0}$
$lnfrac{160}{390-T_0}=ln(frac{200}{390-T_0})^2$
$frac{160}{390-T_0}=(frac{200}{390-T_0})^2$
$390-T_0=frac{200^2}{160}=250$
$T_0=140^{circ}F$
With this value we can find $k=ln(1.25)$
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
$endgroup$
add a comment |
$begingroup$
Newton's law serves equally for cooling or warming situations:
$frac{dT}{dt}=k(T_e-T)$
The general solution is $T(t)=T_e+(T_0-T_e)e^{-kt}$; where $T_0=T(0)$
is the initial reading we are going to find
In this case $T_e=390^{circ}F$ is the oven temperature; so we can write
$T(t)=390+(T_0-390)e^{-kt}$
Now, we have recorded two temperatures at some given moments:
$190=390+(T_0-390)e^{-k(1)}$
$230=390+(T_0-390)e^{-k(2)}$
We have to solve these two last equations to find $T_0$ and $k$
$(T_0-390)e^{-k(1)}=-200$
$(T_0-390)e^{-k(2)}=-160$
$e^{-k}=frac{200}{390-T_0}$
$e^{-2k}=frac{160}{390-T_0}$
$-k=lnfrac{200}{390-T_0}$
$-2k=lnfrac{160}{390-T_0}$
$lnfrac{160}{390-T_0}=2lnfrac{200}{390-T_0}$
$lnfrac{160}{390-T_0}=ln(frac{200}{390-T_0})^2$
$frac{160}{390-T_0}=(frac{200}{390-T_0})^2$
$390-T_0=frac{200^2}{160}=250$
$T_0=140^{circ}F$
With this value we can find $k=ln(1.25)$
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
$endgroup$
Newton's law serves equally for cooling or warming situations:
$frac{dT}{dt}=k(T_e-T)$
The general solution is $T(t)=T_e+(T_0-T_e)e^{-kt}$; where $T_0=T(0)$
is the initial reading we are going to find
In this case $T_e=390^{circ}F$ is the oven temperature; so we can write
$T(t)=390+(T_0-390)e^{-kt}$
Now, we have recorded two temperatures at some given moments:
$190=390+(T_0-390)e^{-k(1)}$
$230=390+(T_0-390)e^{-k(2)}$
We have to solve these two last equations to find $T_0$ and $k$
$(T_0-390)e^{-k(1)}=-200$
$(T_0-390)e^{-k(2)}=-160$
$e^{-k}=frac{200}{390-T_0}$
$e^{-2k}=frac{160}{390-T_0}$
$-k=lnfrac{200}{390-T_0}$
$-2k=lnfrac{160}{390-T_0}$
$lnfrac{160}{390-T_0}=2lnfrac{200}{390-T_0}$
$lnfrac{160}{390-T_0}=ln(frac{200}{390-T_0})^2$
$frac{160}{390-T_0}=(frac{200}{390-T_0})^2$
$390-T_0=frac{200^2}{160}=250$
$T_0=140^{circ}F$
With this value we can find $k=ln(1.25)$
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
answered Feb 21 '17 at 0:09
J.FrisancoJ.Frisanco
512
512
add a comment |
add a comment |
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$begingroup$
Do you know how to solve the differential equation to get temperature as a function of time? I'm trying to understand which part of this solution you need help with.
$endgroup$
– dshapiro
Mar 24 '15 at 14:33
$begingroup$
One hint is that the value of k remains the same. Equate those values to get T0.
$endgroup$
– Krishna
Mar 24 '15 at 14:39