Heat and Thermodynamics

We know, m = V . $\rho$

where, V = volume and $\rho$ = density.

So, we have

${{dQ} \over {dt}} = {h \over {ms}}(T - {T_0}) = {{h(T - {T_0})} \over {V\,.\,\rho s}}$

Since, h, (T $-$ T₀) and V are constant for both beaker.

$\therefore$ ${{dQ} \over {dt}} \propto {1 \over {\rho s}}$

We have given that $\rho$_A = 8 $\times$ 10² kgm^$-$3, $\rho$_B = 10³ kgm^$-$3, s_A = 2000 J kg^$-$1 K^$-$1 and s_B = 4000 J kg^$-$1 K^$-$1, $\rho$_As_A = 16 $\times$ 10⁵ and $\rho$_Bs_B = 4 $\times$ 10⁶

$\rho$_A < $\rho$_B, s_A < s_B and $\rho$_As_A < $\rho$_Bs_B

$ \Rightarrow {1 \over {{\rho _A}{s_A}}} > {1 \over {{\rho _B}{s_B}}} \Rightarrow {{d{Q_A}} \over {dt}} > {{d{Q_B}} \over {dt}}$

So, for container B, rate of cooling is smaller than the container A. Hence, graph of B lies above the graph of A and it is not a straight line (slope of A is greater than B).

Let x grams of water is evaporated.

According to the principle of calorimetry,

Heat lost by freezing water (that turns into ice) = Heat gained by evaporated water

Given, mass of water = 150 g

$ \Rightarrow (150 - x) \times {10^{ - 3}} \times 3.36 \times {10^5} = x \times {10^{ - 3}} \times 2.10 \times {10^6}$

$ \Rightarrow (150 - x) \times 3.36 = 21x$

$ \Rightarrow x = {{150} \over {7.25}} = 20.6$

$\therefore$ $x \approx 20g$

We know that ${v_{rms}} = \sqrt {{{3{k_B}T} \over m}} $. Given, ${k_B} = 1.4 \times {10^{ - 23}}$ J/K; T = 300 K; m = 7 $\times$ 10^$-$27 kg. Therefore,

${v_{rms}} = \sqrt {{{3 \times 1.4 \times {{10}^{ - 23}} \times 300} \over {7 \times {{10}^{ - 27}}}}} $

$ = \sqrt {{{3 \times 300 \times 14 \times {{10}^{ - 23}}} \over {7 \times 10 \times {{10}^{ - 27}}}}} $

$ = \sqrt {{{{3^2} \times {{10}^2} \times 2 \times {{10}^4}} \over {10}}} = 3 \times 10 \times {10^2} \times \sqrt {{2 \over {10}}} $

$ = 3 \times \sqrt {10} \times {10^2} \times \sqrt 2 = 3 \times \sqrt 2 \times \sqrt 5 \times {10^2} \times \sqrt 2 $

$ = 3 \times 2 \times \sqrt 5 \times {10^2}$

${v_{rms}} = 6 \times {10^2} \times \sqrt 5 = 13.41 \times {10^2}$ or ${v_{rms}} = 1.34 \times {10^3}$

Given that two Carnot engines A and B are operated in series. Efficiency of a Carnot engine $\eta = 1 - {{{T_2}} \over {{T_1}}}$; where T₁ is temperature of source and T₂ is temperature of sink. Therefore,

${\eta _A} = 1 - {{{T_{2A}}} \over {{T_{1A}}}} = 1 - {T \over {600}}$; T is heat rejected by A

and ${\eta _B} = 1 - {{{T_{2B}}} \over {{T_{1B}}}} = 1 - {{100} \over T}$; T is heat rejected by A and received by B

Now, ${{{\eta _B}} \over {{\eta _A}}}{{1 - {{100} \over T}} \over {1 - {T \over {600}}}} = \left( {{{T - 100} \over T}} \right) \times {{600} \over {600 - T}}$

Let T = 350 K, therefore,

${{{\eta _B}} \over {{\eta _A}}} = {{350 - 100} \over {350}} \times {{600} \over {600 - 350}} = {{250} \over {350}} \times {{600} \over {25}} = {{600} \over {350}}$

$ \Rightarrow {{{\eta _B}} \over {{\eta _A}}} = {{12} \over 7}$

In process I, volume is changing. Therefore, it is not isochoric. Therefore 'A' is incorrect.

In process II, q = $\Delta$U + W . $\Delta$U = 0 as temperature is constant. Therefore, q = W. Here W = P(V_f $-$ V_i) is positive therefore q is positive i.e., gas absorbs heat. Therefore 'B' is correct.

For process IV, q = $\Delta$U + W. Here $\Delta$ = 0 and W is negative (volume decreases). Therefore q is negative i.e., gas releases heat. 'C' is correct.

For an isobaric process, V $\propto$ T i.e., we will get a straight inclined line in T-V graph. Therefore I and II are NOT isobaric. 'D' is correct.

Process I is adiabatic, therefore, $\Delta$Q = 0

$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ W = $-$$\Delta$U

Thus, W < 0 since volume of gas is decreasing $\Rightarrow$ $\Delta$U > 0 Therefore, temperature of gas increases and no heat is exchanged between gas and surroundings.

Thus, the correct mapping is P $\to$ 3.

Process II is isobaric, therefore, pressure remains constant

$\Rightarrow$ W = P$\Delta$V = 3P₀[3V₀ $-$ V₀] = 6P₀V₀

Therefore, work done by gas in process II is 6P₀V₀.

Thus, the correct mapping is Q $\to$ 4.

Process III is isochoric, therefore, volume remains constant

$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ $\Delta$Q = $\Delta$U

and work done by gas is zero.

Thus, the correct mapping is R $\to$ 1.

Process IV is isothermal process, therefore temperature remains constant

$\Rightarrow$ $\Delta$Q = $\Delta$U + W $\Rightarrow$ $\Delta$Q = W and $\Delta$ = 0

Thus, the correct mapping is S $\to$ 2.

Therefore, option (C) is correct.