Question 660: Thermodynamics & KTG

Question

A body cools from $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in 5 min . In the next time of ' t ' in, the body continues to cool from $50^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. The total time taken by the body to cool from $80^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$ is [The temperature of the surroundings is $20^{\circ} \mathrm{C}$.]

Accepted Answer

According to Newton's law of cooling,
$$ \frac{	heta_1-	heta_2}{t}=K\left(\frac{	heta_1+	heta_2}{2}-	heta_0ight) $$
where $	heta_0=$ temperature of the surroundings.
Situation I: $\frac{80-50}{5}=K\left(\frac{80+50}{2}-20ight)$
$$ \frac{30}{5}=K(45) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
Situation II: $\frac{50-30}{\mathrm{t}}=\mathrm{K}\left(\frac{50+30}{2}-20ight)$
$$ \frac{20}{t}=K(20) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
Dividing equation (i) by equation (ii), we get
$$ \begin{aligned} & \frac{30}{5} 	imes \frac{t}{20}=\frac{45}{20} \ & t=\frac{45 	imes 5}{30} \ 	herefore \quad & t=7.5 \mathrm{mins} \end{aligned} $$
$	herefore \quad$ Total time taken from $80^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$
$$ =7.5+5=12.5 \mathrm{mins} $$

Answer

10 min

Answer

7.5 min

Answer

15.0 min

Answer

12.5 min

Thermodynamics & KTG

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