As the temperature difference as well as the thermal resistance is same for both the cases, so thermal current will also be same for both the cases.
Convection takes more heat upwards.
Detailed solution not required.
According to Newton's law:
Rate of cooling $\propto$ average temperature difference $\Delta\theta$.
$\lambda_{m} = \frac{b}{T} \implies T = \frac{b}{\lambda_{m}} = \frac{2.93 \times 10^{-3}}{4000 \times 10^{-10}} = 7325~K$
$Q = nC_{v}\Delta T$
$Q = \frac{15}{28} \times \frac{5 \times R}{2} \times (4T - T)$ (where $T = 300~K$)
$Q = 10021.33 \approx 10~kJ$
Detailed solution not required.
According to ideal gas equation:
$PV = nRT$ or $\frac{V}{T} = \frac{nR}{P}$
At constant pressure $\frac{V}{T} = \text{constant}$.
Hence graph (1) is correct.
Detailed solution not required.
$W = \frac{R(T_{i} - T_{t})}{\gamma - 1}$
$6R = \frac{R(T - T_{t})}{\left(\frac{5}{3} - 1\right)}$
$T_{t} = (T - 4)~K$
Due to compression the temperature of the system increases to a very high value. This causes the flow of heat from the system to the surroundings, thus decreasing the temperature. This decrease in temperature results in a decrease in pressure.
Detailed solution not required.
In isothermal process temperature remains constant. i.e., $\Delta T = 0$. Hence according to $C = \frac{Q}{m\Delta T} \implies C_{iso} = \infty$.
In adiabatic process, slope of PV-graph $\frac{dP}{dV} = -\gamma\frac{P}{V} \implies |\text{Slope}| \propto \gamma$.
From the given graph $(\text{Slope})_{2} > (\text{Slope})_{1} \implies \gamma_{2} > \gamma_{1}$.
There 1 should correspond to $O_{2}$ ($\gamma = 1.4$) and 2 should correspond to He ($\gamma = 1.66$).
Detailed solution not required.
Heat gain by water = Heat lost by steam
$20 \times 1 \times (80 - 10) = m \times 540 + m \times 1 \times (100 - 80)$
$\implies 1400 = 560~m \implies m = 2.5~g$
Total mass of water $= 20 + 2.5 = 22.5~g$
Fractional change in its density $= \gamma \times \Delta t$
$= 49 \times 10^{-5} \times 30 = 0.0147$
$\frac{F - 32}{9} = \frac{C}{5} \implies F = \frac{9C}{5} + 32$
Here y-intercept is positive.
$H_{1} = H_{2} + H_{3}$
$\frac{(100 - \theta)}{\left(\frac{1}{3kA}\right)} = \frac{(\theta - 50)}{\left(\frac{1}{2kA}\right)} + \frac{(\theta - 0)}{\left(\frac{1}{kA}\right)}$
or $3(100 - \theta) = 2(\theta - 50) + \theta$
or $\theta = \frac{200}{3}^{\circ}C$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$\rho_{0} + \rho_{t_{2}}gl_{2} = \rho_{0} + \rho_{t_{1}}gl_{1} \implies \rho_{t_{2}}l_{2} = \rho_{t_{1}}l_{1}$
$\frac{\rho_{0}}{1 + \gamma t_{2}}l_{2} = \frac{\rho_{0}}{1 + \gamma t_{1}}l_{1}$
$(1 + \gamma t_{1})l_{2} = (1 + \gamma t_{2})l_{1}$
$\gamma(l_{2}t_{1} - l_{1}t_{2}) = l_{1} - l_{2}$
$\gamma = \frac{l_{1} - l_{2}}{l_{2}t_{1} - l_{1}t_{2}}$
$\Delta Q = \Delta U + \Delta W \because \Delta W = 0 \implies \Delta Q = \Delta U = nC_{v}\Delta T$
$= 2 \times \frac{3R}{2} \times (373 - 273) = 300~R$
$0.8 \times 5 = P \times (3 + 5) \implies P = 0.5~m$
For isothermal process $dU = 0$ and work done $= dW = P(V_{2} - V_{1})$
$V_{2} = \frac{V_{1}}{2} = \frac{V}{2} \therefore dW = -\frac{PV}{2}$
Work done by the gas is negative.
$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma} \implies \frac{P_{2}}{P_{1}} = \left[\frac{V_{1}}{V_{2}}\right]^{\gamma} = \left[\frac{4}{1}\right]^{3/2} = \frac{8}{1}$
For monoatomic gas $P^{\gamma-1}T^{\gamma} = \text{const}$. $P \propto T^{\gamma/(\gamma-1)} \implies \gamma = \frac{5}{3}$
$c = \frac{\gamma}{\gamma-1} = \frac{\frac{5}{3}}{\frac{5}{3}-1} = \frac{5}{2}$
$P_{B} = P$ since process is isobaric.
For isothermal process $P_{1}V_{1} = P_{2}V_{2} \implies P_{C} = 2P$
For adiabatic process $PV^{\gamma} \implies P_{A}V^{3/2} = \text{const}$. $P_{A} = 2\sqrt{2}P$
So $P_{A} : P_{B} : P_{C} = 2\sqrt{2} : 1 : 2$
Detailed solution not required.
$Q = U + W$
Where, $U = $ Internal energy
$\therefore U = Q - W = 100 - 75 = 25~J/s = 25~W$
Heat of combustion $= 4 \times 10^{4}~J/g$
Specific heat of water, $c = 4.2~J~g^{-1}{}^{\circ}C^{-1}$
Mass of flowing water, $m = 3.0~\text{litre/min} = 3000~g/\text{min}$
Total heat used, $\Delta Q = mc~\Delta T = 3000 \times 4.2 \times (77 - 27) = 6.3 \times 10^{5}~J/\text{min}$
Rate of consumption $= \frac{6.3 \times 10^{5}}{4 \times 10^{4}} = 15.75~g/\text{min}$
$\lambda = \frac{1}{\sqrt{2}n\pi d^{2}}$
Detailed solution not required.
Detailed solution not required.
$\frac{P_{1}V_{1}}{RT_{1}} = \frac{P_{2}V_{2}}{RT_{2}} \implies P_{2} = \frac{PV(1.1T)}{(1.05)VT} \Rightarrow 1.05P$
Hence final pressure lies between P and 1.1 P
$\lambda \propto \frac{1}{d^{2}} \implies \frac{\lambda_{1}}{\lambda_{2}} = \left[\frac{d_{2}}{d_{1}}\right]^{2} = \left(\frac{2}{1}\right)^{2} = \frac{4}{1}$
Energy $= 300~J/\text{litre} = 300 \times 10^{3}~J/m^{3}$
$P = \frac{2}{3}E = \frac{2 \times 300 \times 10^{3}}{3} = 2 \times 10^{5}~N/m^{2}$
$\gamma_{mix} = \frac{\frac{\mu_{1}\gamma_{1}}{\gamma_{1}-1} + \frac{\mu_{2}\gamma_{2}}{\gamma_{2}-1}}{\frac{\mu_{1}}{\gamma_{1}-1} + \frac{\mu_{2}}{\gamma_{2}-1}}$
Here $\mu_{1} = \mu_{2} = 1$; $\gamma_{1} = \frac{7}{5}$ and $\gamma_{2} = \frac{5}{3}$
$\gamma_{mix} = \frac{3}{2}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{P_{1}V_{1}}{P_{2}V_{2}}} = \sqrt{\frac{4}{1}} = 2$
For polytropic process $PV^{x} = \text{const.}$ Here $x = -1$.
$C = \frac{R}{\gamma-1} + \frac{R}{1-x}$
$C = \frac{R}{\frac{5}{3}-1} + \frac{R}{1 - (-1)} = \frac{3R}{2} + \frac{R}{2} = 2R$
Detailed solution not required.