Finding the initial temperature using Newton's law












0














I am confused about how to solve the given problem :



A metal bar is heated 100c by a heat source. The heat source is removed when the temperature of the metal bar reached to a plateau. Now the metal bar is placed in a room. The room temperature is 25c. After 15 minutes the bar temperature reached to 90c. What was the initial temperature of the metal bar, assume newton's law of cooling applies.



Normally the initial temperature is given for other problems. it's easy to find out the value of the variables using T(t) = ce^kt+Ta formula. but here How to solve the problem?






calculus integration initial-value-problems







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    0














    I am confused about how to solve the given problem :



    A metal bar is heated 100c by a heat source. The heat source is removed when the temperature of the metal bar reached to a plateau. Now the metal bar is placed in a room. The room temperature is 25c. After 15 minutes the bar temperature reached to 90c. What was the initial temperature of the metal bar, assume newton's law of cooling applies.



    Normally the initial temperature is given for other problems. it's easy to find out the value of the variables using T(t) = ce^kt+Ta formula. but here How to solve the problem?






    calculus integration initial-value-problems







    share|cite|improve this question

























      0












      0








      0







      I am confused about how to solve the given problem :



      A metal bar is heated 100c by a heat source. The heat source is removed when the temperature of the metal bar reached to a plateau. Now the metal bar is placed in a room. The room temperature is 25c. After 15 minutes the bar temperature reached to 90c. What was the initial temperature of the metal bar, assume newton's law of cooling applies.



      Normally the initial temperature is given for other problems. it's easy to find out the value of the variables using T(t) = ce^kt+Ta formula. but here How to solve the problem?






      calculus integration initial-value-problems







      share|cite|improve this question













      I am confused about how to solve the given problem :



      A metal bar is heated 100c by a heat source. The heat source is removed when the temperature of the metal bar reached to a plateau. Now the metal bar is placed in a room. The room temperature is 25c. After 15 minutes the bar temperature reached to 90c. What was the initial temperature of the metal bar, assume newton's law of cooling applies.



      Normally the initial temperature is given for other problems. it's easy to find out the value of the variables using T(t) = ce^kt+Ta formula. but here How to solve the problem?






      calculus integration initial-value-problems




      calculus integration initial-value-problems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 at 18:29









      Cynthia

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          Here you have two sets of data: T(0)=100 and T(15)=90 while $T_a=25$. So you have two equations:$$100=c*e^{k*0}+25 (1)$$$$90=c*e^{15k}+25 (2)$$
          From (1) find c, from (2) find k. Now you have equation in the form $$T=c*e^{k*t}+T_a$$
          with c and k known. Here $T_a$ = 100, because "plateau" happens when $T=T_a = 100$. Maybe this problem lacks info about time of heating. If you'll have time of heating, you'll easily solve it.






          share|cite|improve this answer





















          • Thanks for the reply, I was also thinking the same that T(0) =100 but here I have to find out the initial temp. of the metal bar which is also T(0). That is actually confusing. after find out C and K as you solved how to find the initial temperature? Can you please explain how the time of heating can solve the problem easily?
            – Cynthia
            Nov 26 at 20:44










          • T(0) = 100 for cooling process at room temperature, but T(0) is unknown for heating (two different T(0)). You can name them $T_c (0) = 100$ and $T_h (0) = ?$, if it's still confusing. About solution: since you know c,k, and $T_a$ (note: you should use -c, not +c, because not it's not cooling; it's heating - opposite process), last equation has all needed unknowns except t (time of heating). Just solve it for T. This T is the answer. $$T=c*e^{kt}+T_a$$
            – Kelly Shepphard
            Nov 27 at 10:38












          • thank you so much
            – Cynthia
            Dec 1 at 18:13










          • No problem :) I'm glad that this info was useful for you.
            – Kelly Shepphard
            Dec 1 at 18:25










          • Sorry I just need to clear one thing. you said we should use -c as it is heating. so the last equation should be : T=-c∗e^kt+Ta. right?
            – Cynthia
            Dec 1 at 19:12













          Your Answer





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          Here you have two sets of data: T(0)=100 and T(15)=90 while $T_a=25$. So you have two equations:$$100=c*e^{k*0}+25 (1)$$$$90=c*e^{15k}+25 (2)$$
          From (1) find c, from (2) find k. Now you have equation in the form $$T=c*e^{k*t}+T_a$$
          with c and k known. Here $T_a$ = 100, because "plateau" happens when $T=T_a = 100$. Maybe this problem lacks info about time of heating. If you'll have time of heating, you'll easily solve it.






          share|cite|improve this answer





















          • Thanks for the reply, I was also thinking the same that T(0) =100 but here I have to find out the initial temp. of the metal bar which is also T(0). That is actually confusing. after find out C and K as you solved how to find the initial temperature? Can you please explain how the time of heating can solve the problem easily?
            – Cynthia
            Nov 26 at 20:44










          • T(0) = 100 for cooling process at room temperature, but T(0) is unknown for heating (two different T(0)). You can name them $T_c (0) = 100$ and $T_h (0) = ?$, if it's still confusing. About solution: since you know c,k, and $T_a$ (note: you should use -c, not +c, because not it's not cooling; it's heating - opposite process), last equation has all needed unknowns except t (time of heating). Just solve it for T. This T is the answer. $$T=c*e^{kt}+T_a$$
            – Kelly Shepphard
            Nov 27 at 10:38












          • thank you so much
            – Cynthia
            Dec 1 at 18:13










          • No problem :) I'm glad that this info was useful for you.
            – Kelly Shepphard
            Dec 1 at 18:25










          • Sorry I just need to clear one thing. you said we should use -c as it is heating. so the last equation should be : T=-c∗e^kt+Ta. right?
            – Cynthia
            Dec 1 at 19:12


















          0














          Here you have two sets of data: T(0)=100 and T(15)=90 while $T_a=25$. So you have two equations:$$100=c*e^{k*0}+25 (1)$$$$90=c*e^{15k}+25 (2)$$
          From (1) find c, from (2) find k. Now you have equation in the form $$T=c*e^{k*t}+T_a$$
          with c and k known. Here $T_a$ = 100, because "plateau" happens when $T=T_a = 100$. Maybe this problem lacks info about time of heating. If you'll have time of heating, you'll easily solve it.






          share|cite|improve this answer





















          • Thanks for the reply, I was also thinking the same that T(0) =100 but here I have to find out the initial temp. of the metal bar which is also T(0). That is actually confusing. after find out C and K as you solved how to find the initial temperature? Can you please explain how the time of heating can solve the problem easily?
            – Cynthia
            Nov 26 at 20:44










          • T(0) = 100 for cooling process at room temperature, but T(0) is unknown for heating (two different T(0)). You can name them $T_c (0) = 100$ and $T_h (0) = ?$, if it's still confusing. About solution: since you know c,k, and $T_a$ (note: you should use -c, not +c, because not it's not cooling; it's heating - opposite process), last equation has all needed unknowns except t (time of heating). Just solve it for T. This T is the answer. $$T=c*e^{kt}+T_a$$
            – Kelly Shepphard
            Nov 27 at 10:38












          • thank you so much
            – Cynthia
            Dec 1 at 18:13










          • No problem :) I'm glad that this info was useful for you.
            – Kelly Shepphard
            Dec 1 at 18:25










          • Sorry I just need to clear one thing. you said we should use -c as it is heating. so the last equation should be : T=-c∗e^kt+Ta. right?
            – Cynthia
            Dec 1 at 19:12
















          0












          0








          0






          Here you have two sets of data: T(0)=100 and T(15)=90 while $T_a=25$. So you have two equations:$$100=c*e^{k*0}+25 (1)$$$$90=c*e^{15k}+25 (2)$$
          From (1) find c, from (2) find k. Now you have equation in the form $$T=c*e^{k*t}+T_a$$
          with c and k known. Here $T_a$ = 100, because "plateau" happens when $T=T_a = 100$. Maybe this problem lacks info about time of heating. If you'll have time of heating, you'll easily solve it.






          share|cite|improve this answer












          Here you have two sets of data: T(0)=100 and T(15)=90 while $T_a=25$. So you have two equations:$$100=c*e^{k*0}+25 (1)$$$$90=c*e^{15k}+25 (2)$$
          From (1) find c, from (2) find k. Now you have equation in the form $$T=c*e^{k*t}+T_a$$
          with c and k known. Here $T_a$ = 100, because "plateau" happens when $T=T_a = 100$. Maybe this problem lacks info about time of heating. If you'll have time of heating, you'll easily solve it.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 19:47









          Kelly Shepphard

          2298




          2298












          • Thanks for the reply, I was also thinking the same that T(0) =100 but here I have to find out the initial temp. of the metal bar which is also T(0). That is actually confusing. after find out C and K as you solved how to find the initial temperature? Can you please explain how the time of heating can solve the problem easily?
            – Cynthia
            Nov 26 at 20:44










          • T(0) = 100 for cooling process at room temperature, but T(0) is unknown for heating (two different T(0)). You can name them $T_c (0) = 100$ and $T_h (0) = ?$, if it's still confusing. About solution: since you know c,k, and $T_a$ (note: you should use -c, not +c, because not it's not cooling; it's heating - opposite process), last equation has all needed unknowns except t (time of heating). Just solve it for T. This T is the answer. $$T=c*e^{kt}+T_a$$
            – Kelly Shepphard
            Nov 27 at 10:38












          • thank you so much
            – Cynthia
            Dec 1 at 18:13










          • No problem :) I'm glad that this info was useful for you.
            – Kelly Shepphard
            Dec 1 at 18:25










          • Sorry I just need to clear one thing. you said we should use -c as it is heating. so the last equation should be : T=-c∗e^kt+Ta. right?
            – Cynthia
            Dec 1 at 19:12




















          • Thanks for the reply, I was also thinking the same that T(0) =100 but here I have to find out the initial temp. of the metal bar which is also T(0). That is actually confusing. after find out C and K as you solved how to find the initial temperature? Can you please explain how the time of heating can solve the problem easily?
            – Cynthia
            Nov 26 at 20:44










          • T(0) = 100 for cooling process at room temperature, but T(0) is unknown for heating (two different T(0)). You can name them $T_c (0) = 100$ and $T_h (0) = ?$, if it's still confusing. About solution: since you know c,k, and $T_a$ (note: you should use -c, not +c, because not it's not cooling; it's heating - opposite process), last equation has all needed unknowns except t (time of heating). Just solve it for T. This T is the answer. $$T=c*e^{kt}+T_a$$
            – Kelly Shepphard
            Nov 27 at 10:38












          • thank you so much
            – Cynthia
            Dec 1 at 18:13










          • No problem :) I'm glad that this info was useful for you.
            – Kelly Shepphard
            Dec 1 at 18:25










          • Sorry I just need to clear one thing. you said we should use -c as it is heating. so the last equation should be : T=-c∗e^kt+Ta. right?
            – Cynthia
            Dec 1 at 19:12


















          Thanks for the reply, I was also thinking the same that T(0) =100 but here I have to find out the initial temp. of the metal bar which is also T(0). That is actually confusing. after find out C and K as you solved how to find the initial temperature? Can you please explain how the time of heating can solve the problem easily?
          – Cynthia
          Nov 26 at 20:44




          Thanks for the reply, I was also thinking the same that T(0) =100 but here I have to find out the initial temp. of the metal bar which is also T(0). That is actually confusing. after find out C and K as you solved how to find the initial temperature? Can you please explain how the time of heating can solve the problem easily?
          – Cynthia
          Nov 26 at 20:44












          T(0) = 100 for cooling process at room temperature, but T(0) is unknown for heating (two different T(0)). You can name them $T_c (0) = 100$ and $T_h (0) = ?$, if it's still confusing. About solution: since you know c,k, and $T_a$ (note: you should use -c, not +c, because not it's not cooling; it's heating - opposite process), last equation has all needed unknowns except t (time of heating). Just solve it for T. This T is the answer. $$T=c*e^{kt}+T_a$$
          – Kelly Shepphard
          Nov 27 at 10:38






          T(0) = 100 for cooling process at room temperature, but T(0) is unknown for heating (two different T(0)). You can name them $T_c (0) = 100$ and $T_h (0) = ?$, if it's still confusing. About solution: since you know c,k, and $T_a$ (note: you should use -c, not +c, because not it's not cooling; it's heating - opposite process), last equation has all needed unknowns except t (time of heating). Just solve it for T. This T is the answer. $$T=c*e^{kt}+T_a$$
          – Kelly Shepphard
          Nov 27 at 10:38














          thank you so much
          – Cynthia
          Dec 1 at 18:13




          thank you so much
          – Cynthia
          Dec 1 at 18:13












          No problem :) I'm glad that this info was useful for you.
          – Kelly Shepphard
          Dec 1 at 18:25




          No problem :) I'm glad that this info was useful for you.
          – Kelly Shepphard
          Dec 1 at 18:25












          Sorry I just need to clear one thing. you said we should use -c as it is heating. so the last equation should be : T=-c∗e^kt+Ta. right?
          – Cynthia
          Dec 1 at 19:12






          Sorry I just need to clear one thing. you said we should use -c as it is heating. so the last equation should be : T=-c∗e^kt+Ta. right?
          – Cynthia
          Dec 1 at 19:12




















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