Class 11th

Heat is supplied to the system at a rate of 100W

Hence, heat supplied, Q = 100 J/s

The system performs at the rate of 75 J/s

Hence, work done, W = 75 J/s

From the 1st law of Thermodynamics, we have Q = U + W, where U is the internal energy

U = Q – W = 100 – 75 = 25 J/s = 25 W

Therefore the internal energy of the given electric heater increases at a rate of 25 W.

Work done by the steam engine per minute, W = 5.4 $*{10}^8$ J

Heat supplied by the boiler, H = 3.6 $*{10}^9$ J

Efficiency of the engine,  $\eta$ = $\frac{Outputenergy}{Inputenergy}$ = $\frac{5.4\mathrm{}*{10}^8}{3.6\mathrm{}*{10}^9}$ = 0.15

Amount of heat wasted = Input energy – Output energy

= 3.6 $*{10}^9-$ 5.4 $*{10}^8=3.06*{10}^9$ J

The work done, W = 22.3 J

Being an adiabatic process,  $?$ Q = 0

$?$ W = -22.3 J : since the work is done on the system

From the 1st law of thermodynamics, we know $?$ Q = $?U+$ $?$ W, where $?U$ is the change of internal energy of the gas

$?$ U = 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

$?$ Q = 9.35 cal = 9.35 $*4.19$ J = 39.1765 J

Heat absorbed $?$ Q = $?U+$ $?$ W

$?$ W = $?$ Q - $?U$ = 39.1765 – 22.3 = 16.8765 J

Therefore, work done by the system is 16.8765 J

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus the process is 'Adiabatic'.

Let the initial and final pressure inside the cylinder be $P_1$ & $P_2$ and volume be $V_1$ & $V_2$ .

Ratio of specific heat, $\gamma$ = 1.4

For an adiabatic process, we know $P_1{V}_1^{\gamma }$ = $P_2{V}_2^{\gamma }$

It is given $V_2=$ $\frac{V_1}{2}$

Hence $P_1{V}_1^{\gamma }$ = $P_2{\frac{V_1}{2}}^{\gamma }$ or $\frac{P_2}{P_1}$ = ( $V_1^{\gamma })/$ ( ${\frac{V_1}{2}}^{\gamma })$ = $2^{\gamma }$ = $2^{1.4}$ = 2.639

Hence the pressure increases by a factor of 2.639

Mass of Nitrogen, m = 2.0 * ${10}^{-2}$ kg = 20 g

Rise in temperature,  $?T$ = 45 °C

Molecular mass of $N_2$ . M = 28

Universal gas constant, R = 8.3 J mol–1 K–1

Number of moles, n = $\frac{m}{M}$ = $\frac{20}{28}$ = 0.714

Molar specific heat at constant pressure for nitrogen,  $C_p$ = $\frac{7}{2}$ R = 29.05 J/mol/K

The total amount of heat to be supplied is given by the relation

$?$ Q = n $C_p?T$ = 0.714 $*29.05*45$ = 933.38 J

Initial temperature,  $T_1$ = 27 °C

Final temperature,  $T_2$ = 77 °C

Rise in temperature,  $?T=T_2-T_1=50?$

Heat of combustion = 4.0 * ${10}^4$ J/g

Specific heat of water, c = 4.21 J/g/ $?$

Mass of flowing water, m = 3 lit/min = 3000 g/min

Total heat used,  $?Q=mc?T$ = 3000 $*4.21*50$ = 6.315 $*{10}^5$ J/min

Rate of consumption = $\frac{6.315\mathrm{}*{10}^5}{4.0\mathrm{}*\mathrm{}{10}^4}$ g/min = 15.79 g/min

The displacement equation for an oscillating mass is given by : x = Acos ( $\omega t+\theta )$ , where

A = the amplitude

x = the displacement

$\theta =\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{s}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$

Velocity, V = $\frac{dx}{dt}$ = -A $\omega \mathrm{s}\mathrm{i}\mathrm{n}?(\omega$ t + $\theta )$

At t = 0, x = $x_0$ , $x_0=A\cos ?\theta$ = $x_0$ ….(i)

And $\frac{dx}{dt}$ = $-v_0$ = A $\omega \sin ?\theta$ …….(ii)

Squaring and adding, we get

$A^2\left({cos}^2\theta +{sin}^2\theta \right)=x_0^2+\left(\frac{v_0^2}{{\omega }^2}\right)$ , A = $\sqrt{x_0^2+({\frac{v_0}{\omega })}^2}$

Amplitude = 5 cm = 0.05 m

Time period, T = 0.2 s

(a) For displacement x = 5 cm = 0.05m

Acceleration is given by a = ${-\omega }^{2}x$ = $-({\frac{2\pi }{T})}^{2}x$ = $-({\frac{2\pi }{0.2})}^2*0.05$ = $-$ 5 ${\pi }^2$ m/s

Velocity is given by V = $\omega \sqrt{A^2-x^2}$ = $\frac{2\pi }{T}\sqrt{{0.05}^2-{0.05}^2}$ = 0

(b) For displacement x = 3 cm = 0.03m

Acceleration is given by a = ${-\omega }^{2}x$ = $-({\frac{2\pi }{T})}^{2}x$ = $-({\frac{2\pi }{0.2})}^2*0.03$ = $-$ 3 ${\pi }^2$ m/s

Velocity is given by V = $\omega \sqrt{A^2-x^2}$ = $\frac{2\pi }{T}\sqrt{{0.05}^2-{0.03}^2}$ = 0.4 $\pi m/s$

(c) For displacement x = 0 cm = 0 m

Acceleration is given by a = ${-\omega }^{2}x$ = $-({\frac{2\pi }{T})}^{2}x$ = $-({\frac{2\pi }{0.2})}^2*0$ = $0$ m/s

Velocity is given by V = $\omega \sqrt{A^2-x^2}$ = $\frac{2\pi }{T}\sqrt{{0.05}^{2}0}$ = 0.5 $\pi m/s$