Thermodynamics Practice Questions

$E=\sigma A T^4 ; A \propto R^2$ $\therefore E \propto R^2 T^4$ $\therefore \;\frac{E_2}{E_1}=\;\frac{R_2^2 T_2^4}{R_1^2 T_1^4}$ $\Rightarrow \;\frac{E_2}{E_1}$ $=\;\frac{(2 R)^2(2 T)^4}{R^2 T^4}$ $=64$

$\;\frac{n_1+n_2}{\gamma -1}=\;\frac{n_1}{\gamma _1-1}+\;\frac{n_2}{\gamma _2-1}$ $\Rightarrow \;\frac{1+1}{\gamma -1}$ $=\;\frac{1}{\frac{5}{3}-1}+\;\frac{1}{\frac{7}{5}-1}$ $\Rightarrow \gamma =\;\frac{3}{2}$

Internal energy and entropy are state function, they do not depend upon path taken.

Work is a path function. The remaining three parameters are state function.

This is a statement of second law of thermodynamics

$η=1-\;\frac{T_2}{T_1}$ $=1-\;\frac{(273+27)}{(273+627)}$ $=1-\;\frac{300}{900}$ $=1-\;\frac{1}{3}=\;\frac{2}{3}$But $η=\;\frac{W}{Q}$ $\therefore \;\frac{W}{Q}=\;\frac{2}{3} \Rightarrow W$ $=\;\frac{2}{3} \times Q$ $=\;\frac{2}{3} \times 3 \times 10^6$ $=2 \times 10^6 {\text{cal}}$ $=2 \times 10^6 \times 4.2 {\text{J}}$ $=8.4 \times 10^6 {\text{J}}$

$P \propto T^3 \Rightarrow P T^{-3}=\;\text{ constant }\; . . . . .(i)$But for an adiabatic process, the pressure temperature relationship is given by$P^{1-\gamma } T^\gamma =\;\text{ constant }\; \Rightarrow P T^{\;\frac{\gamma }{1-7}}=\;\text{ constant. }\;$From (i) and (ii) $\;\frac{\gamma }{1-\gamma }=-3 \Rightarrow \gamma =-3+3 \gamma \Rightarrow \gamma =\;\frac{3}{2}$

Wein's law correctly explanations the spectrum

$η=1-\;\frac{T_2}{T_1}$ For $η=1$ or $100 \%, T_2=0 K$. The temperature of $0 K$ (absolute zero) can not be obtained.

Black board paint is quite approximately equal to black bodies.

Heat required for raising the temperature of the whole body by $1^{\circ} {\text{C}}$ is called the thermal capacity of the body.

Since pressure and volume are not changing, so temperature remains same.

The energy radiated per second is given by $E=e \sigma T^4 A$ For same material $e$ is same. $\sigma$ is stefan's constant $\therefore \;\frac{E_1}{E_2}=\;\frac{T_1^4 A_1}{T_2^4 A_2}=\;\frac{T_1^4 4 \pi r_1^2}{T_2^4 4 \pi r_2^2}$ $=\;\frac{(4000)^4 \times 1^2}{(2000)^4 \times 4^2}=\;\frac{1}{1}$

When water is cooled to form ice, energy is released from water in the form of heat. As energy is equivalent to mass therefore when water is cooled to ice, its mass decreases.

All reversible engines working for the same temperature of source and sink have same efficiencies. If the temperatures are different, the efficiency is different.

If $n_1$ moles of adiabatic exponent $\gamma _1$ is mixed with $n_2$ moles of adiabatic exponent $\gamma _2$ then the adiabatic component of the resulting mixture is given by $\;\frac{n_1+n_2}{\gamma -1}=\;\frac{n_1}{\gamma _1-1}+\;\frac{n_2}{\gamma _2-1}$ $\;\frac{1+1}{\gamma -1}=\;\frac{1}{\frac{7}{5}-1}+\;\frac{1}{\frac{1}{3}-1}$ $\therefore \;\frac{2}{\gamma -1}=\;\frac{5}{2}+\;\frac{3}{2}=4$ $\therefore 2=4 \gamma -4 \Rightarrow \gamma =\;\frac{6}{4}=\;\frac{3}{2}$

Pyrometer is used to detect infra-red radiation.

$v_{r m s}=\sqrt{\;\frac{R T}{M}}$ For $v_{r m s}$ to be equal $\;\frac{T_{H_2}}{M_{H_2}}=\;\frac{T_{O_2}}{M_{O_2}}$ Here $M_{H_2}=2 ; M_{O_2}=32 ;$ $T_{O_2}=47+273=320 K$ $\therefore \;\frac{T_{H_2}}{2}=\;\frac{320}{32}$ $\Rightarrow T_{H_2}=20 K$

$P_1 V=n_1 R T$ $P_2 V=n_2 R T$ $\Rightarrow (100 {\text{kPa}}) {\text{V}}= ( {\text{n}}_1+ {\text{n}}_2) {\text{RT}}$ $\Rightarrow n_1+n_2=\;{(100 {\text{kPa}}) (2000 {\text{cm}}^3) }/{8.3 \times 300}$Also, $n_1 \times 2+n_2 \times 32=0.76$Solving (1) and (2),${\text{n}}_1=0.06$ ${\text{n}}_2=0.02$ $\Rightarrow \;\frac{n_1}{n_2}=3$

The amount of heat energy required to raise the temperature of a substance can be calculated as:${\text{Q}}= {\text{m}} \times {\text{c}} \times \Delta {\text{T}}$where $Q$ is the heat energy required, ${\text{m}}$ is the mass of the substance, ${\text{c}}$ is its specific heat capacity, and $\Delta T$ is the change in temperature.The time required to heat a substance can be calculated as :${\text{t}}=\;\frac{Q}{P}$where $t$ is the time required, and $P$ is the power of the heating device.The actual power output of the heating device can be calculated as:$P_{\;\text{actual }\;}=P_{\;\text{input }\;} \times \;\text{ efficiency }\;$where $P_{-}$input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.Substituting the given values:${\text{Q}}=2 {\text{kg}} \times 4200 {\text{J}} / {\text{kg}} / {\text{K}} \times (60-10)=2 {\text{kg}} \times 4200 {\text{J}} / {\text{kg}} / {\text{K}} \times 50 {\text{K}}=4200 \times 50 \times 2 {\text{J}}=420,000 {\text{J}}$ $P_{\;\text{input }\;}=2000 {\text{W}}=2000 {\text{J}} / {\ s}$ ${\text{P}}_{\;\text{actual }\;}=2000 \times 0.7=1400 {\text{J}} / {\ s}$ ${\text{t}}=\;{Q}/{P_{\;\text{actual }\;}}=\;\frac{420,000}{1400} {\text{J}} / {\ s}=300 {\ s}$So, the time required to heat $2 {\text{kg}}$ of water from $10^{\circ} {\text{C}}$ to $60^{\circ} {\text{C}}$ is approximately $300 {\ s}$.