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.
enter image description here



However, the given solution actually places a modulus around (25-$theta$), like so.
enter image description here



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 differential-equations physics







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  • Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
    – Simply Beautiful Art
    Dec 22 '16 at 14:01










  • There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
    – David Quinn
    Dec 22 '16 at 14:08














up vote
0
down vote

favorite












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.
enter image description here



However, the given solution actually places a modulus around (25-$theta$), like so.
enter image description here



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 differential-equations physics







share|cite|improve this question





















  • Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
    – Simply Beautiful Art
    Dec 22 '16 at 14:01










  • There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
    – David Quinn
    Dec 22 '16 at 14:08












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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.
enter image description here



However, the given solution actually places a modulus around (25-$theta$), like so.
enter image description here



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 differential-equations physics







share|cite|improve this question













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.
enter image description here



However, the given solution actually places a modulus around (25-$theta$), like so.
enter image description here



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 differential-equations physics




calculus differential-equations physics






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asked Dec 22 '16 at 13:59









Charlz97

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  • Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
    – Simply Beautiful Art
    Dec 22 '16 at 14:01










  • There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
    – David Quinn
    Dec 22 '16 at 14:08
















  • Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
    – Simply Beautiful Art
    Dec 22 '16 at 14:01










  • There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
    – David Quinn
    Dec 22 '16 at 14:08















Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
– Simply Beautiful Art
Dec 22 '16 at 14:01




Haha, well, I mean, sometimes I get unrealistic physics problem for homework.
– Simply Beautiful Art
Dec 22 '16 at 14:01












There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
– David Quinn
Dec 22 '16 at 14:08




There should be a modulus sign when you integrate this, but since $theta<25$ it is irrelevant here
– David Quinn
Dec 22 '16 at 14:08










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What about $$ fracdthetadt=k(25 - theta)$$
$$ fracdtheta25-theta=k cdot dt$$
$$ int_T_0^T fracdtheta25 - theta = int_t_0^t k cdot dtprime$$



$$ -ln left( 25 - theta right) big|_T_0^T = k cdot left( t - t_0 right)$$
$$ ln left( 25 - theta right) big|_T_0^T = - k cdot left( t - t_0 right)$$



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.
$$ ln left( frac25 -T25 - T_0 right)= -kt $$
$$ frac25-T25-T_0 = e^-kt $$
$$ T = 25 + left( T_0 - 25 right) cdot e^-kt $$



We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.






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    What about $$ fracdthetadt=k(25 - theta)$$
    $$ fracdtheta25-theta=k cdot dt$$
    $$ int_T_0^T fracdtheta25 - theta = int_t_0^t k cdot dtprime$$



    $$ -ln left( 25 - theta right) big|_T_0^T = k cdot left( t - t_0 right)$$
    $$ ln left( 25 - theta right) big|_T_0^T = - k cdot left( t - t_0 right)$$



    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.
    $$ ln left( frac25 -T25 - T_0 right)= -kt $$
    $$ frac25-T25-T_0 = e^-kt $$
    $$ T = 25 + left( T_0 - 25 right) cdot e^-kt $$



    We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.






    share|cite|improve this answer


























      up vote
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      down vote













      What about $$ fracdthetadt=k(25 - theta)$$
      $$ fracdtheta25-theta=k cdot dt$$
      $$ int_T_0^T fracdtheta25 - theta = int_t_0^t k cdot dtprime$$



      $$ -ln left( 25 - theta right) big|_T_0^T = k cdot left( t - t_0 right)$$
      $$ ln left( 25 - theta right) big|_T_0^T = - k cdot left( t - t_0 right)$$



      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.
      $$ ln left( frac25 -T25 - T_0 right)= -kt $$
      $$ frac25-T25-T_0 = e^-kt $$
      $$ T = 25 + left( T_0 - 25 right) cdot e^-kt $$



      We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.






      share|cite|improve this answer
























        up vote
        0
        down vote










        up vote
        0
        down vote









        What about $$ fracdthetadt=k(25 - theta)$$
        $$ fracdtheta25-theta=k cdot dt$$
        $$ int_T_0^T fracdtheta25 - theta = int_t_0^t k cdot dtprime$$



        $$ -ln left( 25 - theta right) big|_T_0^T = k cdot left( t - t_0 right)$$
        $$ ln left( 25 - theta right) big|_T_0^T = - k cdot left( t - t_0 right)$$



        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.
        $$ ln left( frac25 -T25 - T_0 right)= -kt $$
        $$ frac25-T25-T_0 = e^-kt $$
        $$ T = 25 + left( T_0 - 25 right) cdot e^-kt $$



        We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.






        share|cite|improve this answer














        What about $$ fracdthetadt=k(25 - theta)$$
        $$ fracdtheta25-theta=k cdot dt$$
        $$ int_T_0^T fracdtheta25 - theta = int_t_0^t k cdot dtprime$$



        $$ -ln left( 25 - theta right) big|_T_0^T = k cdot left( t - t_0 right)$$
        $$ ln left( 25 - theta right) big|_T_0^T = - k cdot left( t - t_0 right)$$



        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.
        $$ ln left( frac25 -T25 - T_0 right)= -kt $$
        $$ frac25-T25-T_0 = e^-kt $$
        $$ T = 25 + left( T_0 - 25 right) cdot e^-kt $$



        We then have $T(0) = T_0$ and $T( infty ) = 25$ as required.







        share|cite|improve this answer














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        edited Dec 22 '16 at 15:02

























        answered Dec 22 '16 at 14:41









        UweM.

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