Question

A hot body placed in a surrounding of temperature ${\theta }_0$ obeys Newton's law of cooling $\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$ Its temperature at t = 0 is θ1. The specific heat capacity of the body is s and its mass is m. Find (i) the maximum heat that the body can lose and (ii) the time starting from t = 0 in which it will lose 90$\%$ of this maximum heat.

• A. (i) ms $\frac{({\theta }_1-{\theta }_0)}{2}$, (ii) $\frac{ln10}{K}$
• B. (i) ms $\frac{({\theta }_1-{\theta }_0)}{2}$, (ii)$\frac{ln10}{2K}$
• C. (i) ms $({\theta }_1-{\theta }_0)$, (ii) $\frac{ln10}{K}$
• D. (i) ms $({\theta }_1-{\theta }_0)$, (ii)$\frac{ln10}{2K}$

Answer 1

The correct option is C

(i) ms $({\theta }_1-{\theta }_0)$, (ii) $\frac{ln10}{K}$

$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$ ....(i)

Also $\Delta Q=ms\Delta \theta$ ....(ii)

Now, the maximum heat the body can lose corresponds to the lowest temperature it can fall to, starting from θ1. And this minimum temperature is the surrounding temperature, that is

${\theta }_0({\theta }_1>{\theta }_0)$.

$\therefore Q_{max}=ms\Delta {\theta }_{max}$ (from ii)

= ms $({\theta }_1-{\theta }_0)$

Now, the temperature at which $\Delta Q=0.9\Delta Q_{max}$ is given by $\Delta Q=0.9\Delta Q_{max}$

$\Rightarrow \Delta \theta =0.9\Delta {\theta }_{max}$

$\Rightarrow {\theta }_1-\theta =0.9({\theta }_1-{\theta }_0)$

$\Rightarrow \theta =0.1{\theta }_1+0.9{\theta }_0$ ....(iii)

Now integrating (i) by using (ii)for limits,

${\int }_{{\theta }_1}^{\theta }\frac{d\theta }{\theta -{\theta }_0}=-k_0^t\int dt$

$ln\left(\frac{\theta -{\theta }_0}{{\theta }_1-{\theta }_0}\right)=-kt$

$\Rightarrow ln\left(\frac{0.1{\theta }_1+0.9{\theta }_0-{\theta }_0}{{\theta }_1-{\theta }_0}\right)=-kt$

$\Rightarrow ln\left(\right(0.1\left)\left(\frac{{\theta }_1-{\theta }_0}{{\theta }_1-{\theta }_0}\right)\right)=-kt$

$\Rightarrow ln{10}^{-1}=-kt$

$\Rightarrow t=\frac{ln10}{k}$