Thermodynamics Test

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Thermodynamics Test - 36

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Weekly Quiz Competition

Question 1
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One end of a $$0.25\ m$$ long metal bar is in steam and the other end is in contact with ice. If $$12\ g$$ of ice melts per minute, what is the thermal conductivity of the metal? Given cross-section of the bar $$= 5\times 10^{-4} m^{2}$$ and latent heat of ice is $$80\ cal/g$$.

A
$$80\ cal/s-m-^{\circ}C$$

B
$$90\ cal/s-m-^{\circ}C$$

C
$$70\ cal/s-m-^{\circ}C$$

D
$$60\ cal/s-m-^{\circ}C$$

Solution
Question 2
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"Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement of the consequence of

A
second law of thermodynamics

B
conservation of momentum

C
conservation of mass

D
first law of thermodynamics

Solution

Hint: The second law of thermodynamics dictates that an isolated system’s entropy will not decrease with time. And first law states that energy cannot be created not be destroyed and that it is simply transferred from one form to the other

Step 1: Explanation:

The Clausius statement on Second law of thermodynamics states that, “Heat cannot flow from a cold body to a hot body without the performance of work by some external agency.” This is in direct correlation to the second law.

$$\textbf{Hence, the correct option is (A)}$$
Question 3
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Select the incorrect option about spontaneous exothermic reaction.

A
$$\Delta S_{Surr} = + ve$$

B
$$\Delta S_{Overall} = + ve$$

C
$$\Delta S_{System} = - ve$$

D
$$\Delta S_{System}= + ve$$ always

Solution

Spontaneous reactions will -ve $$\Delta$$ G [Gibbs free energy]
And exothermic reaction means -ve $$\Delta$$H
From Formula $$\Delta G=\Delta H-\Delta T$$ S
To maintain -$$\Delta$$G overall $$\Delta$$S should be +ve so option B is correct.
As by the exothermic process we are decreasing the entropy (energy) of a system by increasing entropy of surroundings.
Hence $$\Delta S_{surr}$$ = +ve and $$\Delta S_{system}$$ may =-ve are correct.
The last one $$\Delta S_{system}$$ = +ve always is incorrect by considering the above three statements.
Question 4
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Two bars of same length and same cross -sectional area but of different thermal conductivities $$K_1$$ and $$K_2$$ are joined end to end as shown in the figure .One end of the compound bar is at temperature $$T_1$$ and the opposite end at temperature $$T_2$$ $$(where T_1 > T_2)$$.
The temperature of the junction is

A
$$\dfrac{K_1T_1+K_2T_2}{K_1+K_2}$$

B
$$\dfrac{K_1T_2+K_2T_1}{K_1+K_2}$$

C
$$\dfrac{K_1(T_1+T_2)}{K_2}$$

D
$$\dfrac{K_2(T_1+T_2)}{K_1}$$

Solution

Let $$L$$ and $$A$$ be length and area of cross-section of each bar respectively.

$$\therefore $$ Heat current through the bar 1 is
$$H_1=\dfrac{K_1A(T_1-T_0)}{L}$$

At steady state,$$H_1=H_2$$

$$\therefore \dfrac{K_1A(T_1-T_0)}{L}=\dfrac{K_2A(T_0-T_2)}{L}$$

$$K_1(T_1-T_0)=K_2(T_0-T_2)$$

$$K_1T_1-K_1T_0=K_2T_0-K_2T_2$$

$$K_1T_0+K_2T_0=K_1T_1+K_2T_2$$

$$T_0(K_1+K_2)=K_1T_1+K_2T_2$$

$$T_0=\dfrac{K_1T_1+K_2T_2}{(K_1+K_2)}$$
Question 5
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The internal energy of a system remains constant when it undergoes

A
Cyclic process

B
An isothermal process

C
Any process in which the heat given out by the system is equal to work done on the system

D
All of the above

Solution
Question 6
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Which of the following options is correct regarding spontaneity of a process occurring on a system in which only pressure-volume? work is involved and S, G, Cl, H, V, and P have usual meaning as in thermodynamics?

A
$$(dG)_{U, V} < 0, (dS)_{T, V} > 0$$

B
$$(dH)_{S, V} < 0, (dG)_{T, P} < 0$$

C
$$(dU)_{S, V} < 0, (dG)_{T, V} < 0$$

D
$$(dS)_{H, P} > 0, (dG)_{T, P} < 0$$

Solution

For the reaction to be spontenaous $$\Delta G$$ that is gibb's free enrgy should be negative and entropy $$(\Delta S)$$ should be positive.
For reaction to be spontaneous reaction $$(\Delta H)$$ should also be negative i.e exothermic reaction.
Mathematically
$$(dS)_{H.P}>0$$
$$(dG)_{T,G}<0$$
Question 7
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A certain mass of gas is taken from an initial thermodynamic state A to another state B by process I and II. In process I the gas does 5 joules of work and absorbs 4 joules of heat energy. In process II, the gas absorbs 5 joules of heat. The work done by the gas in process II (see figure) is

A
+6 joules

B
-6 joules

C
+4 joules

D
-4 joules

Solution

Internal energy doesnt depend on path.
$$\therefore Q=U+W$$
in path $$I$$
$$Q=u, w=+5$$
$$U=-1\ J$$
in path $$II$$
$$Q=5, U=-1$$
$$\therefore W=5-(-1)=6\ J$$
Question 8
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If Q increases then:

A
$$\Delta G$$ increases and the reaction becomes more spontaneous.

B
$$\Delta G$$ increases and the reaction becomes less spontaneous.

C
$$\Delta G$$ decreases and the reaction becomes more spontaneous.

D
$$\Delta G$$ decreases and the reaction becomes less spontaneous.

Solution

$$ Q = \Delta u+w $$
$$ \Delta G = \Delta H-T\Delta S $$
considering work done
to be constant
$$ Q \uparrow \Delta u\uparrow $$
$$ \Delta G \uparrow $$
hence reactioon becomes spontaneous.
Question 9
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The internal energy $$U$$ is a unique function of any state because change in $$U$$.

A
Does not depends upon path

B
Depends upon path

C
Corresponds to adiabatic process

D
Corresponds to an isothermal process

Solution

Internal energy of a body depends upon the final and initial temperature of the body. It does not depend upon the path an the process the body has undergoes
So, option $$(A)$$
Question 10
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Magnitude of Seebeck emf between the junctions does not depend on

A
thermocouple

B
temperature of cold junction

C
temperature of hot function

D
neutral temperature

Solution

The Seeback effect is the direct conversion of heat into electricity in which potential difference is needed to calculate the current density so the potential of two junctions must be known. It has nothing to do with neutral temperature.