An object cools from 100^{circ} C to 40^{circ | vedclass

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(D) According to Newton's law of cooling, the rate of change of temperature is given by $\frac{d	heta}{dt} = -k(	heta - 	heta_0)$, where $	heta_0$

Vedclass · Accepted Answer

$16.3$

Vedclass · Answer

(D) According to Newton's law of cooling, the rate of change of temperature is given by $\frac{d	heta}{dt} = -k(	heta - 	heta_0)$, where $	heta_0$ is the surrounding temperature.Integrating this, we get $\ln\left(\frac{	heta_1 - 	heta_0}{	heta_2 - 	heta_0}ight) = kt$.For the first case: $	heta_1 = 100^{\circ} C$, $	heta_2 = 40^{\circ} C$, $	heta_0 = 10^{\circ} C$, $t = 10$ min.$kt_1 = \ln\left(\frac{100-10}{40-10}ight) = \ln\left(\frac{90}{30}ight) = \ln 3 = 1.1$.So, $k(10) = 1.1 \Rightarrow k = 0.11 	ext{ min}^{-1}$.For the second case: $	heta_1 = 70^{\circ} C$, $	heta_2 = 20^{\circ} C$, $	heta_0 = 10^{\circ} C$.$kt_2 = \ln\left(\frac{70-10}{20-10}ight) = \ln\left(\frac{60}{10}ight) = \ln 6 = 1.8$.Substituting $k = 0.11$: $0.11 	imes t_2 = 1.8$.$t_2 = \frac{1.8}{0.11} \approx 16.36 	ext{ min}$.Thus, the time taken is approximately $16.3$ min.

An object cools from $100^{\circ} C$ to $40^{\circ} C$ in $10$ minutes, when the surrounding temperature is $10^{\circ} C$. Then the time taken by the object to cool from $70^{\circ} C$ to $20^{\circ} C$ is
$ [\text{Take } \ln 2=0.7, \ln 3=1.1, \ln 6=1.8 ]$ (in $min$)

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An object cools from $100^{\circ} C$ to $40^{\circ} C$ in $10$ minutes, when the surrounding temperature is $10^{\circ} C$. Then the time taken by the object to cool from $70^{\circ} C$ to $20^{\circ} C$ is $ [\text{Take } \ln 2=0.7, \ln 3=1.1, \ln 6=1.8 ]$ (in $min$)