The rate at which the metal cools in moving a | vedclass

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$(A)$ According to Newton's law of cooling,the rate of change of temperature is given by $\frac{dT}{dt} = -k(T - T_m)$,where $T_m = 290 \ K$ is the su

Vedclass · Accepted Answer

$40 \ 	ext{min}$

Vedclass · Answer

$(A)$ According to Newton's law of cooling,the rate of change of temperature is given by $\frac{dT}{dt} = -k(T - T_m)$,where $T_m = 290 \ K$ is the surrounding temperature.Integrating the equation: $\int \frac{dT}{T - 290} = \int -k \ dt \Rightarrow \ln(T - 290) = -kt + C$.At $t = 0$,$T = 370 \ K$: $\ln(370 - 290) = C \Rightarrow C = \ln(80)$.So,$\ln(T - 290) = -kt + \ln(80) \Rightarrow \ln\left(\frac{T - 290}{80}ight) = -kt$.At $t = 10 \ 	ext{min}$,$T = 330 \ K$: $\ln\left(\frac{330 - 290}{80}ight) = -10k \Rightarrow \ln\left(\frac{40}{80}ight) = -10k \Rightarrow \ln\left(\frac{1}{2}ight) = -10k \Rightarrow -\ln(2) = -10k \Rightarrow k = \frac{\ln(2)}{10}$.Now,for $T = 295 \ K$: $\ln\left(\frac{295 - 290}{80}ight) = -kt \Rightarrow \ln\left(\frac{5}{80}ight) = -\left(\frac{\ln(2)}{10}ight)t$.$\ln\left(\frac{1}{16}ight) = -\frac{\ln(2)}{10}t \Rightarrow -\ln(16) = -\frac{\ln(2)}{10}t$.$-4 \ln(2) = -\frac{\ln(2)}{10}t \Rightarrow t = 40 \ 	ext{minutes}$.

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