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  • Help me answer: Temperature of body incrases at the rate proportional to differnce between its temperature and temperature of surrounding,then diffrential equation will be.

Temperature of body incrases at the rate proportional to differnce between its temperature and temperature of surrounding,then diffrential equation will be.

  • Option 1)

    \frac{dT}{dt}=K(T-T_{0})

  • Option 2)

    \frac{dT}{dt}=K

  • Option 3)

    \frac{dT}{dt}=K(T^{2}-T_{0})

  • Option 4)

    None of these

 

Answers (1)

As we have learned

Temperature Problems -

\frac{dT}{dt}= -k\left ( T-T_{m} \right )

- wherein

K is Proportionality constant

T = Temperature of body

T_{m}=  Temperature of Surrounding

 

 

\frac{dT}{dt}\propto (T-T_0) \Rightarrow \frac{dT}{dt} = K(T-T_0)


Option 1)

\frac{dT}{dt}=K(T-T_{0})

Option 2)

\frac{dT}{dt}=K

Option 3)

\frac{dT}{dt}=K(T^{2}-T_{0})

Option 4)

None of these

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subam

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