Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

(c) According to Newton's law of cooling
Rate of cooling ∝ Mean temperature difference

$\Rightarrow \frac{\mathrm{Fall} \mathrm{in} \mathrm{temperature}}{\mathrm{Time}}\propto \left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0\right)$

$$\begin{gathered} \because {\left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}\right)}_1>{\left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}\right)}_2>{\left(\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}\right)}_3 \\ \Rightarrow {\mathrm{T}}_1<{\mathrm{T}}_2<{\mathrm{T}}_3 \end{gathered}$$

(a) Initially at t = 0
Rate of cooling (R) ∝ Fall in temperature of body $\left(\mathrm{\theta }-{\mathrm{\theta }}_0\right)$

$\Rightarrow \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{{\mathrm{\theta }}_1-{\mathrm{\theta }}_0}{{\mathrm{\theta }}_2-{\mathrm{\theta }}_0}=\frac{100-40}{80-40}=\frac{3}{2}$

(a) For small difference of temperature, it is the special case of Stefan’s law.

(c) According to Newton's law of cooling

$\frac{∆\mathrm{T}}{∆\mathrm{t}}=\frac{4{\mathrm{\sigma eAT}}_0^3}{\mathrm{ms}}\left[\frac{{\mathrm{T}}_1+{\mathrm{T}}_2}{2}-{\mathrm{T}}_0\right]$

where T₁ : initial temperature

          T₂ : final temperature

For container containing 350 g water

$\frac{70-60}{180}=\frac{4{\mathrm{\sigma eAT}}_0^3}{{\mathrm{m}}_1{\mathrm{s}}_1}\left[\frac{70+60}{2}-{\mathrm{T}}_0\right] ... \left(\mathrm{i}\right)$

For container containing 300 g liquid,

$\frac{70-60}{95}=\frac{4{\mathrm{\sigma eAT}}_0^3}{{\mathrm{m}}_2{\mathrm{s}}_2}\left[\frac{70+60}{2}-{\mathrm{T}}_0\right] ... \left(\mathrm{ii}\right)$

Dividing (i) by (ii)

$$\begin{gathered} \frac{95}{180}=\frac{{\mathrm{m}}_2{\mathrm{s}}_2}{{\mathrm{m}}_1{\mathrm{s}}_1} \\ 95\left[\left(350+10\right)\times 1\right]=180\left[\left(300\times \mathrm{C}\right)+10\times 1\right] \end{gathered}$$

C = 0.6 cal/g^oC

(b) During clear nights object on surface of earth radiate out heat and temperature falls. Hence option (a) is wrong.
The total energy radiated by a body per unit time per unit area $\mathrm{E}\propto {\mathrm{T}}^4$. Hence option (c) is wrong.
Energy radiated per second is given by $\frac{\mathrm{Q}}{\mathrm{t}}={\mathrm{PA\epsilon \sigma T}}^4$
$\because {\mathrm{P}}_1={\mathrm{P}}_2$, hence option (d) is wrong.
Newton's law is an approximate form of Stefan's law of radiation and works well for natural convection. Hence option (b) is correct.

(d) $\frac{\mathrm{d\theta }}{\mathrm{dt}}=\frac{\mathrm{\sigma A}}{\mathrm{mc}}\left({\mathrm{T}}^4-{\mathrm{T}}_0^4\right)$ . If the liquids put in exactly similar calorimeters and identical surrounding then we can consider ${\mathrm{T}}_0$ and A constant then   $\frac{\mathrm{d\theta }}{\mathrm{dt}}\propto \frac{\left({\mathrm{T}}^4-{\mathrm{T}}_0^4\right)}{\mathrm{mc}}$  ……(i)
If we consider that equal masses of liquid (m) are taken at the same temperature then $\frac{\mathrm{d\theta }}{\mathrm{dt}}\propto \frac{1}{\mathrm{c}}$ 
So for same rate of cooling c should be equal which is not possible because liquids are of different nature. Again from equation (i)
$\frac{\mathrm{d\theta }}{\mathrm{dt}}\propto \frac{\left({\mathrm{T}}^4-{\mathrm{T}}_0^4\right)}{\mathrm{mc}}\Rightarrow \frac{\mathrm{d\theta }}{\mathrm{dt}}\propto \frac{\left({\mathrm{T}}^4-{\mathrm{T}}_0^4\right)}{\mathrm{V\rho c}}$
Now if we consider that equal volume of liquid (V) are taken at the same temperature then  $\frac{\mathrm{d\theta }}{\mathrm{dt}}\propto \frac{1}{\mathrm{\rho c}}$.
So for same rate of cooling ,multiplication of $\mathrm{\rho }\times \mathrm{c}$ for two liquid of different nature to be same is possible. So option (d) may be correct.

(a) $\frac{365-361}{2}=\mathrm{k}\left[\frac{365+361}{2}-293\right]=70 \mathrm{k}\Rightarrow \mathrm{k}=\frac{1}{35}$

Again $\frac{344-342}{\mathrm{t}}=\frac{1}{35}\left[\frac{344-342}{2}-293\right]=\frac{10}{7}$

$\Rightarrow \mathrm{t}=\frac{14}{10}\min =\frac{14}{10}\times 60=84 \sec .$

(a) I holds good only for small temperature difference between the body and the surroundings i.e. less than 10$°$C.

(b) In first case $\frac{50-40}{10}=\mathrm{K}\left[\frac{{\displaystyle 50+40}}{{\displaystyle 2}}-20\right]$           ….(i)
In second case $\frac{40-{\mathrm{\theta }}_2}{10}=\mathrm{K}\left[\frac{40+{\mathrm{\theta }}_2}{2}-20\right]$            ….(ii)
By solving ${\mathrm{\theta }}_2=33.3°\mathrm{C}$.

(c) $$\begin{gathered} \frac{90-60}{5}=\mathrm{K}\left(\frac{90+60}{2}-20\right)\Rightarrow 6=\mathrm{k}\times 55\Rightarrow \mathrm{k}=\frac{6}{55} \\ \mathrm{And}, \frac{60-30}{\mathrm{t}}=\frac{6}{55}\left(\frac{60+30}{2}-20\right)\Rightarrow \mathrm{t}=11 \mathrm{minute} \end{gathered}$$