Heat and Thermodynamics

T is final temperature, when equilibrium is reached

Heat gained = Heat lost

$n{C_p}\Delta T = n{C_v}\Delta T$

$2 \times {{7R} \over 2} \times (T - 400) = 2 \times {{3R} \over 2} \times (700 - T)$

$7T - 2800 = 2100 - 3T$

$10T = 4900$

$T = 490K$

Heat lost by monoatomic gas = Heat gained by diatomic gas

$2 \times {5 \over 2}R \times (700 - T) = 2 \times {5 \over 2}R(T - 400)$

$3500 - 5T = 7T - 2800$

$12T = 6300$ or, $T = 525K$

Work done by monoatomic gas:

$ = {n_1}R\Delta T$ ($\because$ PV = nRT)

$ = 2 \times R \times (700 - 525)$ ($\Delta$T is negative)

$ = - 350R$

Work done by diatomic gas $ = {n_2}R\Delta T$

$ = 2R(525 - 400) = 250R$. (There is gain in temperature)

The net work done = $-$350 + 250 = $-$100R and $\Delta$T is positive.

According to definition of specific heat capacity,

$C = {{\Delta Q} \over {m\Delta T}}$

$ \Rightarrow \Delta Q = mC\Delta T$ $\therefore$ ${{\Delta Q} \over {\Delta t}} = mC{{\Delta T} \over {\Delta t}}$

Rate of heat absorbed $R = {{\Delta Q} \over {\Delta t}}$

$ \Rightarrow {{\Delta Q} \over {\Delta t}} \propto C$

(a) In 0-100 K,

C increases with T but not linearly. So R increases but not linearly.

(b) As $\Delta Q = mC\Delta T$

$Q = m\int {C\Delta T} $

= m area under C-T curve

From the graph it is clear that area under C-T is more in 400-500 K than in 0-100 K.

Therefore, heat absorbed in 0-100 K is less than in 400-500 K.

(c) In 400-500 K,

C remains constant so there is no change in R.

(d) In 200-300 K,

C increases so R increases.

From the given figure, it can be concluded that process FG is isothermal and process FH is adiabatic.

${\gamma _{mono}} = 1 + {2 \over f} = {5 \over 3}$

${P_0}V_G^{5/3} = 32{P_0}V_0^{5/3}$

${V_G} = {(32)^{3/5}}{V_0} = 8{V_0}$

$\Delta {W_{GE}} = {P_0}({V_0} - 32{V_0}) = - 31{P_0}{V_0}$

Thus, the correct mapping is (P) $\to$ (4).

$\Delta {W_{GH}} = {P_0}(8{V_0} - 32{V_0}) = - 24{P_0}{V_0}$

Thus, the correct mapping is (Q) $\to$ (3).

$\Delta {W_{FH}} = {{{P_0}(8{V_0}) - 32{P_0}{V_0}} \over {1 - (5/3)}} = {{ - 24{P_0}{V_0}} \over { - (2/3)}} = 36{P_0}{V_0}$

Thus the correct mapping is (R) $\to$ (2).

$\Delta {W_{FG}} = 32R{T_0}\ln 32 = 32R{T_0}\ln {2^5} = 160R{T_0}\ln 2 = 160{P_0}{V_0}\ln 2$

Thus, the correct mapping is (S) $\to$ (1).

To determine the ratio of the densities of the two non-reactive monoatomic ideal gases, we can use the ideal gas law and the given information. The ideal gas law is given by:

$ PV = nRT $

Where:

• P is the pressure
• V is the volume
• n is the number of moles
• R is the gas constant
• T is the temperature

Given that the gases are monoatomic and non-reactive, their molecular masses are directly proportional to their atomic masses. Let the atomic masses of the two gases be $M_1$ and $M_2$, with $M_1 : M_2 = 2 : 3$.

The density ($\rho$) of a gas is given by the formula:

$ \rho = \frac{m}{V} $

Where $m$ is the mass and $V$ is the volume. Using the ideal gas law, we can relate density to pressure and temperature.

From the ideal gas law:

$ PV = nRT $

Since $ n = \frac{m}{M} $, where $M$ is the molar mass, we have:

$ P = \frac{mRT}{MV} $

Rearranging to solve for density ($\rho\ = \frac{m}{V}$):

$ \rho = \frac{PM}{RT} $

Thus, the density of a gas is directly proportional to its pressure and molar mass, and inversely proportional to the temperature. Given that the ratio of their partial pressures $P_1 : P_2 = 4 : 3$, we can write:

$ \rho_1 = \frac{P_1 M_1}{RT} $

$ \rho_2 = \frac{P_2 M_2}{RT} $

Taking the ratio of the densities:

$ \frac{\rho_1}{\rho_2} = \frac{\left(\frac{P_1 M_1}{RT}\right)}{\left(\frac{P_2 M_2}{RT}\right)} = \frac{P_1 M_1}{P_2 M_2} $

Substituting the given ratios, $ P_1 : P_2 = 4 : 3 $ and $ M_1 : M_2 = 2 : 3 $, into the equation, we get:

$ \frac{\rho_1}{\rho_2} = \frac{4 \times 2}{3 \times 3} = \frac{8}{9} $

Therefore, the ratio of their densities is $ \frac{8}{9} $, which corresponds to option D. Consequently, the answer is:

Option D: 8 : 9

Let L and A be length and area of cross-section of each block.

$\therefore$ Thermal resistance of block 1 is

$\therefore$ ${R_1} = {L \over {\kappa A}}$

Thermal resistance of block 2 is

${R_2} = {L \over {2\kappa A}}$

In configuration I, two blocks are connected in series. So, their equivalent thermal resistance is

${R_S} = {R_1} + {R_2} = {L \over {\kappa A}} + {L \over {2\kappa A}} = {3 \over 2}{L \over {\kappa A}}$ ...... (i)

$\therefore$ Rate of heat flow in configuration I is

$\therefore$ ${Q \over t} = {{{T_1} - {T_2}} \over {{R_s}}}$ ..... (ii)

In configuration II, two blocks are connected in parallel. So, their equivalent thermal resistance is

${1 \over {{R_P}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} = {1 \over {{L \over {\kappa A}}}} + {1 \over {{L \over {2\kappa A}}}} = {{3\kappa A} \over L}$ ...... (iii)

${R_P} = {1 \over 3}{L \over {\kappa A}}$

$\therefore$ Rate of same amount of heat flow in configuration II is

$\therefore$ ${Q \over {t'}} = {{{T_1} - {T_2}} \over {{R_P}}}$ ...... (iv)

Divide (ii) by (iv), we get

${{t'} \over t} = {{{R_P}} \over {{R_S}}} = {1 \over 3}{L \over {\kappa A}} \times {{2\kappa A} \over {3L}} = {2 \over 9}$ (Using (i) and (iii))

$t' = {2 \over 9}t = {2 \over 9} \times 9s = 2s$

Here's a breakdown of how to solve this problem:

Understanding the Concepts

• Ideal Gas: An ideal gas is a theoretical gas that follows the ideal gas law perfectly. The ideal gas law is a relationship between pressure ($P$), volume ($V$), temperature ($T$), and the number of moles ($n$) of a gas:

$PV = nRT$, where $R$ is the ideal gas constant.

• Heat Capacity: Heat capacity is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius (or 1 Kelvin). For an ideal gas, we can use the specific heat capacity at constant volume ($C_v$) or at constant pressure ($C_p$).


• Constant Volume Process: In a constant volume process, the volume of the gas remains constant. The heat required to raise the temperature is given by:

$Q = nC_v\Delta T$

• Constant Pressure Process: In a constant pressure process, the pressure of the gas remains constant. The heat required to raise the temperature is given by:

$Q = nC_p\Delta T$

• Molar Heat Capacities of Helium: Helium is a monatomic gas. For monatomic gases, we have:


• $C_v = \frac{3}{2}R$


• $C_p = \frac{5}{2}R$

Solving the Problem

1. Identify the Process: Since the balloon is fully expandable, the pressure remains constant. This is a constant pressure process.


2. Calculate the Heat: Use the formula for heat at constant pressure:

$Q = nC_p\Delta T$

1. Substitute the Values:

• $n = 2 \text{ moles}$


• $C_p = \frac{5}{2}R = \frac{5}{2}(8.31 \text{ J/mol.K})$


• $\Delta T = 35^\circ C - 30^\circ C = 5 K$

1. Solve for Q:

$Q = (2 \text{ moles}) \times \left(\frac{5}{2} \times 8.31 \text{ J/mol.K}\right) \times (5 K) $

$Q = 207.75 \text{ J}$

Answer:

The amount of heat required to raise the temperature of the helium gas is approximately 208 J. Therefore, the correct answer is Option D.

The root mean square (rms) speed of a gas is given by the formula :

$v_{rms} = \sqrt{\frac{3kT}{m}}$

where :

• $k$ is the Boltzmann constant
• $T$ is the temperature in Kelvin
• $m$ is the mass of one molecule of the gas

Given that the atomic mass of helium is 4 amu and that of argon is 40 amu, we can find the rms speed ratio by comparing the rms speeds of helium and argon, as temperature $T$ and Boltzmann constant $k$ are the same for both gases :

$\frac{v_{rms} \left( \text{helium} \right)}{v_{rms} \left( \text{argon} \right)} = \sqrt{\frac{m_{\text{argon}}}{m_{\text{helium}}}}$

Substituting the masses :

$m_{\text{helium}} = 4 \text{ amu}$

$m_{\text{argon}} = 40 \text{ amu}$

we get :

$\frac{v_{rms} \left( \text{helium} \right)}{v_{rms} \left( \text{argon} \right)} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$

Thus, the ratio of the rms speeds of helium to argon is approximately 3.16.

The correct answer is :

Option D : 3.16

Let the steady state temperature of the middle plate to T₀. In the steady state, the heat radiated per unit time by the middle plate is equal to the heat received per unit time by it. The middle plate radiates heat from both the surfaces and receives the heat radiated by the first and the third plates.

Stefan-Boltzmann law gives the rate of heat loss and heat gain by the middle plate as

$d{Q_{out}}/dt = \sigma AT_0^4 + \sigma AT_0^4$,

$d{Q_{in}}/dt = \sigma A{(2T)^4} + \sigma A{(3T)^4}$.

The steady state condition, $d{Q_{out}}/dt = d{Q_{in}}/dt$

$\sigma A{(2T)^4} + \sigma A{(3T)^4} = \sigma 2A{({T_0})^4}$

$16{T^4} + 81{T^4} = 2{({T_0})^4}$

$97{T^4} = 2{({T_0})^4}$

${({T_0})^4} = {{97} \over 2}{T^4} = {\left( {{{97} \over 2}} \right)^{1/4}}T$

Initially

V₁ = 5.6 l, T₁ = 273 K, P₁ = 1 atm,

$\gamma = {5 \over 3}$ (For monoatomic gas)

The number of moles of gas is $n = {{5.6l} \over {22.4l}} = {1 \over 4}$

Finally (after adiabatic compression)

V₂ = 0.7 l

For adiabatic compression ${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$

$\therefore$ ${T_2} = {T_1}{\left( {{{{V_1}} \over {{V_2}}}} \right)^{\gamma - 1}} = {T_1}{\left( {{{5.6} \over {0.7}}} \right)^{{5 \over 3} - 1}} = {T_1}{(8)^{2/3}} = 4{T_1}$

Work done during an adiabatic process is

$W = {{nR[{T_1} - {T_2}]} \over {(\gamma - 1)}} = {{{1 \over 4}R[{T_1} - 4{T_1}]} \over {\left[ {{5 \over 3} - 1} \right]}} = - {9 \over 8}R{T_1}$

Negative sign shows that work is done on the gas.

Process A $\to$ B,

It is a isobaric process

P = constant, V $\propto$ T

$\because$ V_B < V_A $\Rightarrow$ T_B < T_A

$\Delta$U = nC_V$\Delta$T = $-$ve

Hence internal energy decreases.

$\Delta$Q = nC_P$\Delta$T = $-$ve

Hence heat is lost.

$\Delta$W = nR$\Delta$T = $-$ve

Hence, work is done on the gas.

Process, B $\to$ C,

It is a isochoric process.

V = constant, P $\propto$ T

$\because$ P_C < P_B $\Rightarrow$ T_C < T_B

$\Delta$U = nC_V$\Delta$T = $-$ve

Hence, internal energy decreases.

$\Delta$W = 0, $\Delta$Q = $\Delta$U = $-$ve

Hence, heat is lost.

Process C $\to$ D,

It is a isobaric process.

P = constant, V $\propto$ T

$\because$ V_D > V_C $\Rightarrow$ T_D > T_C

$\Delta$U = nC_V$\Delta$T = +ve

Hence internal energy increases.

$\Delta$Q = nC_P$\Delta$T = +ve

Hence, heat is gained.

$\Delta$W = nR$\Delta$T = +ve

Hence, work is done by the gas.

Process, D $\to$ A

According to ideal gas equation

${{{P_A}{V_A}} \over {{T_A}}} = {{{P_D}{V_D}} \over {{T_D}}} \Rightarrow {{(3P)(3V)} \over {{T_A}}} = {{(P)(9V)} \over {{T_D}}} \Rightarrow {T_D} = {T_A}$

Hence, it is a isothermal process.

$\therefore$ $\Delta$U = 0

$\because$ V_A < V_D

$\Delta$W = $-$ve, hence work done is done on the gas.

$\Delta$Q = $\Delta$W = $-$ve, hence heat is lost.

The internal energy of one mole of an ideal gas at temperature T is given by U = 3RT/2. The process AB is isothermal i.e., T_A = T_B = T₀, which makes U_A = U_B.

The work done in the isothermal process AB is

W_AB = nRT ln(V_B/V_A)

= nRT₀ ln(4V₀/V₀) = p₀V₀ ln4.

Information regarding p and T at C can not be obtained from the given graph. Unless it is mentioned that line BC passes through origin or not.

Hence, the correct options are (a) and (b).

Note:

If we assume that the line BC pass through the origin. In the process BC, the slope V/T is constant. Thus, BC is an isobaric process (as V/T = nR/p, by the ideal gas equation). Thus,

p_C = p_B = RT_B / V_B

= RT₀/(4V₀) = (RT₀/V₀)4 = p₀/4.

Apply the ideal gas equation for the states A and C to get

${T_C} = {{{p_C}{V_C}} \over {{p_0}{V_0}}}{T_0} = {{({p_0}/4){V_0}} \over {{p_0}{V_0}}}{T_0} = {T_0}/4$.

We observe the following:

$\bullet$ For the path B $\to$ C $\to$ D, the volume (V) decreases; $\Delta$W < 0.

$\Delta U < 0 \Rightarrow \Delta Q < 0$

$\bullet$ For the process A $\to$ B $\to$ C, $\Delta$W is the area of the semicircle and $\Delta W \ne 0$.

$\bullet$ For the process A $\to$ B (semicircle), it cannot be an isothermal process (whose PV-diagram is a rectangular hyperbola).

$\bullet$ For (clockwise) process A $\to$ B $\to$ C $\to$ D $\to$ A, $\Delta$W is the area enclosed, which is positive.

We know that ${C_P} - {C_V} = R$ is same for all gases. For a diatomic gas, we have

${C_V} = {{5R} \over 2}$

Therefore, ${C_P} = {C_V} + R = {{7R} \over 2}$

For a monoatomic gas, we have

${C_V} = {{3R} \over 2}$

${C_P} = {{5R} \over 2}$

Therefore,

${C_P} + {C_V} = 6R$ (diatomic)

${C_P} + {C_V} = 4R$ (monoatomic)

${{{C_P}} \over {{C_V}}} = {7 \over 5} = 1.4$ (diatomic)

${{{C_P}} \over {{C_V}}} = {5 \over 3} = 1.67$ (monoatomic)

${C_P}\,.\,{C_V} = {{35{R^2}} \over 4}$ (diatomic)

${C_P}\,.\,{C_V} = {{15{R^2}} \over 4}$ (monoatomic)

(A)$\to$(P), (Q), (T)

$\bullet$ Case (P) : $U=\frac{1}{2}CV^2$

$\bullet$ Case (Q) : Since the work is done on the system, the internal energy increases.

$\bullet$ Case (T) : time varying $\overrightarrow B $ creates an induced $\overrightarrow E $ causing current to flow. This dissipates heat in the loop: $H=L^2(Rt)$.

(B)$\to$(Q) : As explained above.

(C)$\to$(S)

$\bullet$ Case (S) : Mass defect is converted into energy which is released.

(D)$\to$(S) : As explained above.

As the ideal gas expands in vacuum, no work is done i.e., w = 0

Container is insulated, so no heat lost or gained i.e., Q = 0.

According to the first law of thermodynamics

$\Delta$U = Q + w

$\therefore$ $\Delta$U = 0

Therefore, no change in the temperature of the gas.

(b) $\to$ ($p, r$) :

Given, PV$^2$ = constant ... (i)

For ideal gas, ${{PV} \over T}$ = constant .... (ii)

From (i) and (ii) V $\mu$ T = constant

As gas expands, its volume increase and temperature decreases.

$\therefore$ (p) is the correct.

Also, Q = nC$\Delta$T .... (i)

Where C = molar specific heat

For a polytropic process,

$C = {C_V} + {R \over {1 - n}}$ and

PV$^n$ = constant

Here, PV$^2$ = constant, where n = 2

$\therefore$ $C = {C_V} + {R \over {1.2}} = {C_V} - R$

For monoatomic gas,

${C_V} = {3 \over 2}R$

$\therefore$ $C = {3 \over 2}R - R = {R \over 2}$

Substituting this value in (1), we get

$Q = n \times {R \over 2} \times \Delta T$

Temperature decreases i.e., $\Delta$T is negative. Therefore, Q is negative. This is turn mean that heat is lost by the gas during the process.

So (r) is the correct option.

(C) $\to$ (P, S)

V$^{1/3}$ $\times$ T = constant

$\Rightarrow$ As the gas expands and volume increases, the temperature decreases. Therefore (P) is the correct option.

In the process, $x = {4 \over 3}$

$\therefore$ $C = {C_V} + {R \over {1 - {4 \over 3}}} = {3 \over 2}R + {{3R} \over { - 1}}$

$ = {3 \over 2}R - 3R = - {{3R} \over 2}$

$\therefore$ $Q = n\left( { - {3 \over 2}R} \right)\Delta t$

As $\Delta t$ = $-$negative ($-$ve), Q = positive (+ve). That means heat is gained by the gas during the process.

(s) is correct.

(d) $\to$ (q, s)

$\Delta T = {{\Delta (PV)} \over {nR}}$

$\Delta (PV)$ = Positive $\therefore$ $\Delta$T = positive

$\therefore$ Temperature increase (Q) is the correct option.

From graph it is clear that, during the process the pressure of the gas increases which shows that the internal energy of the gas has increased. Also, the volume increases which means work is done by the system which needs energy. We conclude from the above fact that gas gains heat during the process.

(S) is correct.

Given, PT$^2$ = constant ..... (i)

For an ideal gas,

$\frac{\mathrm{PV}}{\mathrm{T}}$ = constant ...... (ii)

From above two equation, after eliminating P.

$\frac{\mathrm{V}}{\mathrm{T^3}}$ = Constant

$\Rightarrow$ V = KT$^3$, where $k$ = constant.

$\frac{d\mathrm{V}}{\mathrm{V}}=3\frac{d\mathrm{T}}{\mathrm{T}}$

$ \Rightarrow dV = \left( {{3 \over T}} \right)VdT$ ..... (iii)

Change in volume due to thermal expansion is given by $dV=Vydt$ ..... (iv)

Where, $y$ = coefficient of volume expansion

From equation (iii) and (iv), we have,

$VydT = \left( {{3 \over T}} \right)VdT$

$ \Rightarrow y = {3 \over T}$

Statement 1 :

Total translational kinetic energy $=\frac{3}{2} n \mathrm{RT}$

where,

n = no. of moles

R = gas constant

T = Temperature

but, PV = nRT

$\therefore$ K.E. = 1.5 PV

Statement 2 : Molecules of a gas collide with each other and the velocities of the molecules change due to collision.

But statement 2 is not a correct explanation of statement 1 .

When the piston is pulled out slowly, the pressure drop produced inside the cylinder is almost instantaneously neutralized by the air entering from outside into the cylinder. So, the pressure inside the cylinder is P$_0$ throughout the slow pulling process.

At equilibrium, P = P

$\Rightarrow$ P = P$_0$ + (L$_0$ $-$ H)$\rho$g ..... (i)

$ \Rightarrow {P_0}\pi (\pi {R^2}{L_0}) = P[\pi {R^2}({L_0} - H)]$

$ \Rightarrow P = {{{L_0}{P_0}} \over {{L_0} - H}}$ ..... (ii)

From (i) as (ii)

$ \Rightarrow {{{L_0}{P_0}} \over {{L_0} - H}} = {P_0} + ({L_0} - H)\rho g$

$ \Rightarrow {L_0}{P_0} = {P_0}({L_0} - H) + {({L_0} - H)^2}\rho g$

$ \Rightarrow \rho g{({L_0} - H)^2} + {P_0}({L_0} - H) - {L_0}{P_0} = 0$

Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.

The reason given in question is self explanatory. With an increase in temperature, the entire Plank's curve shifts upwards and hence radiation in any spectrum will increase. Reflected energy per unit time will be zero since black body has zero reflectivity and hence it will remain constant.

The bodies which radiate energy in the form of photons can be determined using Kirchoff's

law of radiation. When these photons reach another surface, they may be absorbed, reflected or transmitted. Since the radiation is continuously falling on the black body, the quantity of radiation absorbed per second will increase.

Therefore, the quantity of radiation absorbed by the black body in unit time will increase, since emissivity = absorptivity, hence the quantity of radiation emitted by black body in unit time will increases, Black body radiates more energy in unit time in the visible spectrum, the reflected energy in unit time by the black body remains same.

Hence, the option (A), (B), (C) and (D) are the correct answer.

$ \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}, \mathrm{~W}=\int \mathrm{P} d \mathrm{~V} $

Along JK : V is constant, P is decreasing and

$\mathrm{P} \propto \mathrm{T}, \mathrm{T}$ should also decrease.

As $\mathrm{PV}=n \mathrm{RT}$.

$ \begin{aligned} \therefore & & \mathrm{W} & =P \Delta V=0 \\ & & \Delta \mathrm{U} & =\Delta \mathrm{Q}<0 \quad \text { (negative) } \end{aligned} $

$\Rightarrow$ Along KL : $\mathrm{P}=$ constant, i.e., isobaric process.

$\therefore n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{T}$ is increasing. So temperature should also increase.

$ \begin{aligned} & \therefore \Delta \mathrm{W}=\mathrm{PdV}>0, \Delta \mathrm{U}=n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{~T}>0 \text { and } \\ & \mathrm{Q}=m c \Delta T>0 . \end{aligned} $

$\therefore$ Volume increases, $\Delta \mathrm{W}>0$

$\Rightarrow$ Along LM : MV = constant, $\Delta \mathrm{V}=0$ [isochoric process]

both P increases and T increases.

$ \therefore W=0, \Delta U>0 \text { and } Q>0 \quad[\Delta Q=\Delta U] $

$\Rightarrow$ Along $\mathrm{MJ}: \mathrm{V}$ is decreasing, $\therefore \Delta \mathrm{W}<0$

T is also decreasing, $\Delta \mathrm{V}<0$ and $\Delta \mathrm{Q}<0$

P increases.

(A) Heat supplied to the cylinder is used up in if raising temperature of cylinder $=m s \Delta \mathrm{T}$

(i) $\mathrm{Q}_{1}=m s \Delta \mathrm{T}=1 \times 400(\mathrm{~T}-20) \text {...(i) }$

S = 400 J kg$^{-10}$C$^{-1}$

(ii) doing work $=\mathrm{P}_{\text {atom }}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

Work done $=\mathrm{P}_{\mathrm{atm}}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$

$\mathrm{Q}_{2}=10^{5} \times\left(\mathrm{V}_{1} \gamma \Delta \mathrm{T}\right)$

$\mathrm{Q}_{2}=10^{5} \times \frac{\text { mass }}{\text { density }} \times \gamma(\mathrm{T}-20)$

$\mathrm{Q}_{2}=10^{5} \times \frac{1}{900} \times 9 \times 10^{-5} \times(\mathrm{T}-20)$

$\mathrm{Q}_{2}=0.001(\mathrm{~T}-20)$ ...... (ii)

$\therefore \quad \mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q}$ Heat supplied.

(Given $\mathrm{Q}=20000 \mathrm{~J}$ )

400.$(T-20)+0.001(T-20)=20000$

$400.001(\mathrm{~T}-20)=2000$

$\mathrm{T}-20=\frac{20000}{400.001}=50$ (approx)

$\mathrm{T}=50+20=70^{\circ} \mathrm{C}$ ...... (iii)

(B) Workdone by cylinder

$\begin{aligned} \mathrm{W} & =\mathrm{Q}_{2}=0.001(70-2 a) \\ & =0.001(50) \\ \mathrm{W} & =0.05 \mathrm{~J} .\end{aligned}$

(C) Change in internal energy

$\begin{aligned} \Delta \mathrm{U} & =20000-0.05 \\ & =19999.95 \mathrm{~J}=1.9 \times 10^{4} \mathrm{~J} \end{aligned}$