Test: MCQs (One or More Correct Option): Heat & Thermodynamics | JEE Advanced - JEE MCQ

∴ The gas is H₂

Heat lost by steam = Heat gained by   (water + calorimeter)

mL + m × c × (100 – 80) = 1.12 × c × (80 – 15)

m [540 + 1 × 20] = 1.12 × 1 × 65

m = 0.13 kg

Total transfer of heat per second through the composite = Heat transfer per second from material with thermal conductivity K₁ + Heat transfer per second from material with thermal conductivity K₂.

(a) For all thermal processes.

(b) According to first law of thermodynamics.

In an adiabatic process ΔQ= 0.

(c) In the isothermal process, ΔT= 0.

(d) In the adiabatic process, ΔQ= 0.

Note : All t hr ee vessels are at same temper ature.
According to Maxwell's distribution of speed, average speed of molecules of a gas

∴ The velocity of oxygen molecules will be same in A as well as C.

The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isothermal line. This is because the work done is the area under the P-V indicator diagram. As shown by the diagram the area under the graph in first diagram will be more than in second diagram. When we extrapolate the graph shown in figure (i). Let P₀ be the intercept on P-axis and V₀ be the intercept on V-axis. The equation of the line AB can be written as

Relation between P and T is the equation of a parabola.

Note :The above equation is of a parabola (between T and V) Differentiating the above equation w.r.t.V, we get

is the value for maxima of temperatureAlso,  P_A V_A = P_B V_B

⇒ TA= TB (From Boyle's law)

⇒ In going from A to B, the temperature of the gas first increase to a maximum and the decreases and
reaches back to the same value.

 Energy emitted per second by body

Energy emitted per second by body

Given that power radiated are equal  

According to Wein's displacement law

Since temperature of A is more therefore

Also according to Wein's displacement law

On solving (i) and (ii), we get

For an isothermal process;  PV = constant On differentiating, we get  ;  PdV + VdP = 0

A is free to move, therefore heat will be supplied at constant pressure 

B is held fixed, therefore heat will be supplied at constant volume.

There is a decrease in volume during melting of an ice slab at 273 K. Therefore, negative work is done by ice-water system on the atmosphere or positive work is done on the ice-water system by the atmosphere. Hence, option (b) is correct.

NOTE : Secondly heat is absorbed during melting (i.e. dQ is positive) and as we have seen, work done by ice-water system is negative (dW is negative.) Therefore, from first law of thermodynamics dU = dQ – dW change in internal energy of ice-water system, dU will be positive or internal energy will increase.

According to Wien's displacement law,   

The wavelength at the peak of the spectrum becomes

NOTE : Thus, the maximum energy is radiated for 10³ nm wavelength. It follows that the energy radiated between 499 nm to 500 nm will be less than that emitted between 999 nm to

 Co-efficient of linear expansion of brass is greater than that of copper i.e., α_B >α_C . Now,,

.

Dividing (i) and (ii)

We know that.

From these expressions, we can conclude that

Also the average kinetic energy of gaseous molecules is

NOTE : The law of equipartition of energy states that 'For a dynamical system in thermal equilibrium, the energy of a system is equally distributed among its various degrees of freedom and the energy associated with each degree of freedom per molecule is

1/2 K.T. Inthis case, O2 and N2 both have two degrees of rotational kinetic energy and since the temperature is also same, the ratio of the average rotational kinetic energy is 1 : 1

Since sun rays fall on the black body, it will absorb more radiation and since, its temperature is constant it will emit more radiation. The temperature will remain same only when energy emitted is equal to energy absorbed.

For monoatomic gas 

For diatomic gas 

In case of an isothermal process we get a rectangular hyperbola in a P-V diagram. Therefore option (a) is wrong.T_D < T_B. Therefore in process B→C→D, ΔU is negative.PV decreases and volume also decreases, therefore W is negative. From first law of thermodynamic, Q is negative i.e., there is a heat loss option (b) is correct.
W_AB > W_BC.
Therefore work done during path A →B→C is positive, option (c) is wrong.
Work done is clockwise cycle in a PV diagram is positive.
Option (d) is correct.

Process A to B As the temperature remains the same, this process is isothermal. Therefore there is no change in the internal energy. Option (a) is correct.

Work done

The process BC is not clear. Therefore no judgement can be made for point C.

It is given that heat Q flows only from left to right through the blocks. Therefore heat flow through A and E slabs are the same.

∴ [a] is correct option

Since heat flow through slabs A and E is same, [b] is not correct.
We know that resistance to heat flow is R =l/KA
Let the width of slabs be Z. Then

Now,ΔT = QR

As R_E is least, ΔT_E is also smallest ie since the resistance to heat flow is least for slab E, the temperature difference across is smallest.

∴  Option (c) is the correct answer.
Also

∴ (d) is the correct option.

We know that dQ = m C dT in the range 0 to 100K From the graph, C increases linearly with temperature therefore the rate at which heat is absorbed varies linearly with temperature. Option (a) is correct As the value of C is greater in the temperature range 400-500K, the heat absorbed in increasing the temperature from 0 - 100K is less than the heat required for increasing the temperature from 400 - 500K option (b) is correct.
From the graph it is clear that the value of C does not change in the temperature range 400-500K, therefore there is no change in the rate of heat absorption in this range.
Option (c) is correct.
As the value of C increases from 200-300K, the rate of heat absorption increases in the range 200-300K. Option (d) is also correct.

∴ options [A], [B] and [C] are correct.

Applying combined gas law

Now change in internal energy

For monoatomic gas f = 3

∴ (b) is the correct option.
Now assuming that the pressure on the piston on the right hand side (not considering the affect of spring) remains the same throughout the motion of the piston then,

where k is spring constant and A = area of piston Energy stored = 1/2 kx²

Now

C is correct option Heat supplied

Q = W + ΔU