Differential equations: temperature change and room temperature
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Problem: A can of orange juice is taken from the fridge and placed in a room where the temperature is a constant $25^circ$C. As the can of orange juice warms up, the rate of increase of its temperature $theta$$^circ$C after time t, where t is measured in minutes, is proportional to the temperature difference (25-$theta$)$^circ$C. Set up a differential equation to represent the above scenario, and solve this differential equation.
The picture shows my initial steps that I took to solve this.
However, the given solution actually places a modulus around (25-$theta$), like so.
I'm puzzled - can (25-θ) actually be less than zero? Because this means that something taken out from the fridge, and then warmed up to room temperature, will then become higher than the room temperature! Or is there something else I'm missing?
calculus ordinary-differential-equations physics
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
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add a comment |
$begingroup$
Problem: A can of orange juice is taken from the fridge and placed in a room where the temperature is a constant $25^circ$C. As the can of orange juice warms up, the rate of increase of its temperature $theta$$^circ$C after time t, where t is measured in minutes, is proportional to the temperature difference (25-$theta$)$^circ$C. Set up a differential equation to represent the above scenario, and solve this differential equation.
The picture shows my initial steps that I took to solve this.
However, the given solution actually places a modulus around (25-$theta$), like so.
I'm puzzled - can (25-θ) actually be less than zero? Because this means that something taken out from the fridge, and then warmed up to room temperature, will then become higher than the room temperature! Or is there something else I'm missing?
calculus ordinary-differential-equations physics
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
$endgroup$
$begingroup$
Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
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– Simply Beautiful Art
Dec 22 '16 at 14:01
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There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
$endgroup$
– David Quinn
Dec 22 '16 at 14:08
add a comment |
$begingroup$
Problem: A can of orange juice is taken from the fridge and placed in a room where the temperature is a constant $25^circ$C. As the can of orange juice warms up, the rate of increase of its temperature $theta$$^circ$C after time t, where t is measured in minutes, is proportional to the temperature difference (25-$theta$)$^circ$C. Set up a differential equation to represent the above scenario, and solve this differential equation.
The picture shows my initial steps that I took to solve this.
However, the given solution actually places a modulus around (25-$theta$), like so.
I'm puzzled - can (25-θ) actually be less than zero? Because this means that something taken out from the fridge, and then warmed up to room temperature, will then become higher than the room temperature! Or is there something else I'm missing?
calculus ordinary-differential-equations physics
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
$endgroup$
Problem: A can of orange juice is taken from the fridge and placed in a room where the temperature is a constant $25^circ$C. As the can of orange juice warms up, the rate of increase of its temperature $theta$$^circ$C after time t, where t is measured in minutes, is proportional to the temperature difference (25-$theta$)$^circ$C. Set up a differential equation to represent the above scenario, and solve this differential equation.
The picture shows my initial steps that I took to solve this.
However, the given solution actually places a modulus around (25-$theta$), like so.
I'm puzzled - can (25-θ) actually be less than zero? Because this means that something taken out from the fridge, and then warmed up to room temperature, will then become higher than the room temperature! Or is there something else I'm missing?
calculus ordinary-differential-equations physics
calculus ordinary-differential-equations physics
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
asked Dec 22 '16 at 13:59
Charlz97Charlz97
12312
12312
$begingroup$
Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
$endgroup$
– Simply Beautiful Art
Dec 22 '16 at 14:01
$begingroup$
There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
$endgroup$
– David Quinn
Dec 22 '16 at 14:08
add a comment |
$begingroup$
Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
$endgroup$
– Simply Beautiful Art
Dec 22 '16 at 14:01
$begingroup$
There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
$endgroup$
– David Quinn
Dec 22 '16 at 14:08
$begingroup$
Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
$endgroup$
– Simply Beautiful Art
Dec 22 '16 at 14:01
$begingroup$
Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
$endgroup$
– Simply Beautiful Art
Dec 22 '16 at 14:01
$begingroup$
There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
$endgroup$
– David Quinn
Dec 22 '16 at 14:08
$begingroup$
There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
$endgroup$
– David Quinn
Dec 22 '16 at 14:08
add a comment |
1 Answer
1
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What about $$ frac{dtheta}{dt}=k(25 - theta)$$
$$ frac{dtheta}{25-theta}=k cdot dt$$
$$ int_{T_0}^T frac{dtheta}{25 - theta} = int_{t_0}^t k cdot dtprime$$
$$ -ln left( 25 - theta right) big|_{T_0}^T = k cdot left( t - t_0 right)$$
$$ ln left( 25 - theta right) big|_{T_0}^T = - k cdot left( t - t_0 right)$$
Let $ t_0 = 0 $ (start time). $T_0$ is the temperature at time 0 which is the temperature of the fridge. The argument $25 - theta $ is always positive.
$$ ln left( frac{25 -T}{25 - T_0} right)= -kt $$
$$ frac{25-T}{25-T_0} = e^{-kt} $$
$$ T = 25 + left( T_0 - 25 right) cdot e^{-kt} $$
We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What about $$ frac{dtheta}{dt}=k(25 - theta)$$
$$ frac{dtheta}{25-theta}=k cdot dt$$
$$ int_{T_0}^T frac{dtheta}{25 - theta} = int_{t_0}^t k cdot dtprime$$
$$ -ln left( 25 - theta right) big|_{T_0}^T = k cdot left( t - t_0 right)$$
$$ ln left( 25 - theta right) big|_{T_0}^T = - k cdot left( t - t_0 right)$$
Let $ t_0 = 0 $ (start time). $T_0$ is the temperature at time 0 which is the temperature of the fridge. The argument $25 - theta $ is always positive.
$$ ln left( frac{25 -T}{25 - T_0} right)= -kt $$
$$ frac{25-T}{25-T_0} = e^{-kt} $$
$$ T = 25 + left( T_0 - 25 right) cdot e^{-kt} $$
We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
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add a comment |
$begingroup$
What about $$ frac{dtheta}{dt}=k(25 - theta)$$
$$ frac{dtheta}{25-theta}=k cdot dt$$
$$ int_{T_0}^T frac{dtheta}{25 - theta} = int_{t_0}^t k cdot dtprime$$
$$ -ln left( 25 - theta right) big|_{T_0}^T = k cdot left( t - t_0 right)$$
$$ ln left( 25 - theta right) big|_{T_0}^T = - k cdot left( t - t_0 right)$$
Let $ t_0 = 0 $ (start time). $T_0$ is the temperature at time 0 which is the temperature of the fridge. The argument $25 - theta $ is always positive.
$$ ln left( frac{25 -T}{25 - T_0} right)= -kt $$
$$ frac{25-T}{25-T_0} = e^{-kt} $$
$$ T = 25 + left( T_0 - 25 right) cdot e^{-kt} $$
We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
$endgroup$
add a comment |
$begingroup$
What about $$ frac{dtheta}{dt}=k(25 - theta)$$
$$ frac{dtheta}{25-theta}=k cdot dt$$
$$ int_{T_0}^T frac{dtheta}{25 - theta} = int_{t_0}^t k cdot dtprime$$
$$ -ln left( 25 - theta right) big|_{T_0}^T = k cdot left( t - t_0 right)$$
$$ ln left( 25 - theta right) big|_{T_0}^T = - k cdot left( t - t_0 right)$$
Let $ t_0 = 0 $ (start time). $T_0$ is the temperature at time 0 which is the temperature of the fridge. The argument $25 - theta $ is always positive.
$$ ln left( frac{25 -T}{25 - T_0} right)= -kt $$
$$ frac{25-T}{25-T_0} = e^{-kt} $$
$$ T = 25 + left( T_0 - 25 right) cdot e^{-kt} $$
We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
$endgroup$
What about $$ frac{dtheta}{dt}=k(25 - theta)$$
$$ frac{dtheta}{25-theta}=k cdot dt$$
$$ int_{T_0}^T frac{dtheta}{25 - theta} = int_{t_0}^t k cdot dtprime$$
$$ -ln left( 25 - theta right) big|_{T_0}^T = k cdot left( t - t_0 right)$$
$$ ln left( 25 - theta right) big|_{T_0}^T = - k cdot left( t - t_0 right)$$
Let $ t_0 = 0 $ (start time). $T_0$ is the temperature at time 0 which is the temperature of the fridge. The argument $25 - theta $ is always positive.
$$ ln left( frac{25 -T}{25 - T_0} right)= -kt $$
$$ frac{25-T}{25-T_0} = e^{-kt} $$
$$ T = 25 + left( T_0 - 25 right) cdot e^{-kt} $$
We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
edited Dec 22 '16 at 15:02
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
answered Dec 22 '16 at 14:41
UweM.UweM.
31516
31516
add a comment |
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$begingroup$
Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
$endgroup$
– Simply Beautiful Art
Dec 22 '16 at 14:01
$begingroup$
There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
$endgroup$
– David Quinn
Dec 22 '16 at 14:08