Chapter 10: Thermal Properties of Matter

Ans: It is known that a body possessing a large reflectivity turns out to be a poor absorber of light radiations. 

A poor absorber would further be a poor emitter of radiation. 

Clearly, a body possessing large reflectivity is a poor emitter.

1. a brass tumbler feels much colder than a wooden tray on a chilly day

Ans: It is known that brass is a good conductor of heat. 

The moment at which one touches a brass tumbler, heat gets conducted from the body to the brass tumbler easily. 

Thus, the temperature of the body lessens to a lower value and one feels cooler.

On the other hand, the moment at which one touches a wooden tray, very little heat is conducted from the body to the wooden tray because wood is a poor conductor of heat. 

Thus, there is only a negligible drop in the temperature of the body and one does not feel cool.

Clearly, a brass tumbler feels colder than a wooden tray on a chilly day.

1. An optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red-hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.

Ans: An optical pyrometer calibrated for an ideal black body radiation provided too low a value for temperature of a red-hot iron piece kept in the open. 

Black body radiation equation is given by: 

$E=\sigma \left( {{T}^{4}}-T_{0}^{4} \right)$ 

Where, 

E = Energy radiation 

T = Temperature of optical pyrometer 

${{T}_{0}}$  = Temperature of open space 

$\sigma =$ Constant

Thus, an increase in the temperature of open space lessens the radiation energy. If the same piece of iron is placed in a furnace, the radiation energy, \[E=\sigma {{T}^{4}}\].

1. the earth without its atmosphere would be inhospitably cold

Ans: It is known that excluding its atmosphere, earth would be inhospitably cold. 

When atmospheric gases are absent, no extra heat would be trapped. 

Clearly, all the heat would be radiated back from the earth’s surface.

1. heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Ans: A heating system on the basis of the circulation of steam would be more efficient in warming a building than that on the basis of the circulation of hot water. 

This can be reasoned as steam contains surplus heat in the form of latent heat $\left( 540cal/g \right)$. 

Ans: With respect to Newton’s law of cooling;

$-\frac{dT}{dt}=K\left( T-{{T}_{0}} \right)$ 

$\Rightarrow \frac{dT}{K\left( T-{{T}_{0}} \right)}=-Kdt$

Where, 

Temperature of the body = T 

Temperature of the surroundings $={{T}_{S}}=20{}^\circ C$ 

K is a constant 

Temperature of the body falls from $80{}^\circ C$ to $50{}^\circ C$ in time, t = 5 min = 300 s 

Integrating equation (a), we get:

$\int\limits_{50}^{80}{\frac{dt}{K\left( T-{{T}_{0}} \right)}=-}\int\limits_{0}^{300}{Kdt}$

$\Rightarrow \left[ {{\log }_{e}}\left( T-{{T}_{0}} \right) \right]_{50}^{80}=-K\left[ t \right]_{0}^{300}$ 

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}\frac{80-20}{50-20}=-300$ 

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}2=-300$ 

$\Rightarrow -\frac{2.3026}{300}{{\log }_{10}}2=K$ 

The temperature of the body falls from $60{}^\circ C$ to $30{}^\circ C$ in time = t

Thus, we get:

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}\frac{60-20}{30-20}=-t$

$\Rightarrow \frac{2.3026}{K}{{\log }_{10}}4=K$

Equating equations (b) and (c), we get:

$\frac{-2.3026}{K}{{\log }_{10}}4=\frac{2.3026}{K}{{\log }_{10}}2$

$\Rightarrow t=300\times 2=600s=10\min $ 

Clearly, the time taken to cool the body from $60{}^\circ C$ to $30{}^\circ C$ is 10 minutes.