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  • Can someone explain Rate at which a substance cools in moving air is propotional to defference between temperature of subtance and that of surrounding. If temperature of surrounding is 290 K and substance cools from 370 K to 330 K in 10 min, then aft

Rate at which a substance cools in moving air is propotional to defference between temperature of subtance and that of surrounding. If temperature of surrounding is 290 K and substance cools from 370 K to 330 K in 10 min, then after what time from initial, temperature will be 295 K?

  • Option 1)

    10 mins

  • Option 2)

    20 mins

  • Option 3)

    30 mins

  • Option 4)

    40 mins

 

Answers (1)

best_answer

As we have learnt,

 

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

 

 Let at time 't', temperature is T.

So,

-\frac{\mathrm{d} T}{\mathrm{d} t} = k(T-290) \Rightarrow \frac{dT}{T-290} = -kdt

On integrating, we get

\ln(t-290) = \kt + c

At t = 0, T = 370 \Rightarrow c = \ln 80

\\*\Rightarrow\ln (t -290) = -kt + \ln(80)

Also, at t = 10, T = 330 \Rightarrow \ln 40 =-10k\ln 80

\\*\Rightarrow 10k = \ln 2 \Rightarrow K = \frac{1}{10}\ln 2 \\*\Rightarrow \ln(T - 290) = \left(-\frac{1}{10} \ln2\right )t + \ln 80

Now when T =295 K then \ln 5 = -\frac{t\ln2}{10} + \ln 80

\Rightarrow \frac{t\ln 2}{10} =\ln 16 \Rightarrow t = 40

 

 

 


Option 1)

10 mins

Option 2)

20 mins

Option 3)

30 mins

Option 4)

40 mins

Posted by

Himanshu

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