Physics - Heat & Thermo Question with Solution | TestHub

According to Newton's law of cooling $\frac{{\mathbf{\theta }}_2-{\mathbf{\theta }}_1}{\mathrm{t}}=\mathrm{K}\left(\frac{{\mathbf{\theta }}_1+{\mathbf{\theta }}_2}{2}-{\mathbf{\theta }}_2\right)$ Since the temperature decreases from $60^{\circ }\mathrm{C}$ to $40^{\circ }\mathrm{C}$ in 7 minutes

$$\frac{60-40}{7}=K\Rightarrow \frac{20}{7}=K(50-10)\Rightarrow K=\frac{1}{14}$$

If the temp. of object becomes ${\mathbf{\theta }}^{\prime }$ in next 7 minutes then $\frac{40-{\mathbf{\theta }}^{\prime }}{7}=\frac{1}{14}\left(\frac{40+{\mathbf{\theta }}^{\prime }}{2}-10\right)$

$$\Rightarrow 40-{\theta }^{\prime }=\frac{1}{4}\left(40+{\theta }^{\prime }-20\right)\Rightarrow 160-4{\theta }^{\prime }=20+{\theta }^{\prime }\Rightarrow 5{\theta }^{\prime }=140\Rightarrow {\theta }^{\prime }=28^{\circ }\mathrm{C}$$

OR According to Newton's law of cooling

$$\begin{array}{rlr}& -\frac{\mathrm{d}\theta }{\mathrm{dt}}=\mathrm{K}\left(\theta -{\theta }_0\right)\text{ or }\mathrm{dt}=-\frac{1}{\mathrm{K}}\frac{\mathrm{d}\theta }{\left(\theta -{\theta }_0\right)}& \therefore {\int }_0^{\mathrm{t}}\mathrm{dt}=-\frac{1}{\mathrm{K}}{\int }_{{\theta }_1}^{{\theta }_2}\frac{\mathrm{d}\theta }{\left(\theta -{\theta }_0\right)}\Rightarrow \mathrm{t}=\frac{1}{\mathrm{K}}{\log }_{\mathrm{e}}\left\{\frac{\left({\theta }_1-{\theta }_0\right)}{\left({\theta }_2-{\theta }_0\right)}\right\}\end{array}$$

As per the question $7=\frac{1}{\mathrm{K}}{\log }_{\mathrm{e}}\left\{\frac{60-10}{40-10}\right\}$

$$\text{ also }7=\frac{1}{K}{\log }_e\left\{\frac{40-10}{\theta -10}\right\}$$

from above equations we have ${\log }_{\mathrm{e}}\left(\frac{50}{30}\right)={\log }_{\mathrm{e}}\left(\frac{30}{\theta -10}\right)$

$$\therefore \frac{5}{3}=\frac{30}{\theta -10}\Rightarrow 5-50=90\Rightarrow \theta =\frac{100}{5}=28^{\circ }\mathrm{C}$$