Miscellaneous Exercises Thermal Physics:-

Problem:

Icebergs in the North Atlantic present hazards to shipping as shown in the below figure, causing the length of shipping routes to increase by about 30% during the iceberg season. Strategies for destroying icebergs include planting explosives, bombing, torpedoing, shelling, ramming, and painting with lampblack. Suppose that direct melting of the iceberg , by placing heat sources in the ice, is tried. How much heat is required to melt 10% of a 210,000 – metric-ton iceberg? (One metric ton = 1000 kg.)

Concept:-

The amount of heat per unit mass that must be transferred to produce a phase change is called the latent L for the process. The total heat transferred in a phase change is then

Q = Lm,

where m is the mass of the sample that changes phase. The heat transferred during melting or freezing is called the heat of fusion.

To obtain the heat that is required to melt 10% of a 210,000-metric-ton iceberg, substitute  210,000-metric-ton for mass of the iceberg m and 333 ×10³ J/kg for latent heat of fusion of water in the equation Q = 10% (Lm),

Q = 10% (Lm)

   = 0.10 (333 ×10³ J/kg) (210,000-metric-ton)

  = 0.10 (333 ×10³ J/kg) (210,000-metric-ton) (10³ kg/1 metric-ton)

  = 0.10(333 ×10³ J/kg) (2.1×10⁸ kg)

  = 7.0×10¹² J

From the above observation we conclude that, the heat that required to melt 10% of a 210,000-metric-ton iceberg would be 7.0×10¹² J.

___________________________________________________________________________

Problem:

For the Carnot cycle shown in below figure, calculate (a) the heat that enters and (b) the work done on the system.

Concept:-

In an isothermal process, the heat Q transfer takes place at constant temperature T. The entropy ΔS of the system is defined as,

ΔS = Q/T

Here Q is the positive heat flow into the system. So the entropy ΔS of the system will increase.

From the equation ΔS = Q/T, the absorbed heat Q will be,

Q = (ΔS) (T)

     = T (S_f – S_i)  

Here S_f is the final entropy of the system and S_i is the initial entropy of he system.

Q + W = 0

Here Q is the heat added to the system and W is the work done.

(a) We have to calculate the heat Q_in that enters into the system.

In the above cycle, the heat only enters along the top path AB.

To obtain the heat Q_in that enters into the system, substitute 400 K for T, 0.6 J/K for S_f and 0.1 J/K for S_i in the equation Q = T (S_f – S_i),

Q_in = T (S_f – S_i)

= (400 K) (0.6 J/K - 0.1 J/K)

= (400 K) (0.5 J/K)

= 200 J

Therefore the heat Q_in that enters into the system would be 200 J.

    = - (Q_in+ Q_out)        (Since, Q = Q_in+ Q_out)

To obtain the work done W on the system, substitute 200 J for Q_in and -125 J for Q_out in the equation W = - (Q_in+ Q_out),

W = - (Q_in+ Q_out)

    = - (200 J+(-125 J) )

    = -75 J 

From the above observation we conclude that, the work done W on the system would be -75 J.

______________________________________________________________________________

Problem:

A Carnot engine works between temperatures T₁ and T₂. It derives a Carnot refrigerator that works between two different temperatures T₃ and T₄ as shown in the below figure. Find the ratio |Q₃| / |Q₁| in terms of the four temperatures.

Concept:-

The efficiency ε of a Carnot engine is equal to the,

ε = 1- T_L/T_H

    = |W|/|Q_H|

Here, T_L is the temperature of sink (lower temperature), T_H is the temperature of source, W is the work done by the engine and Q_H is the heat at higher temperature.

The coefficient of performance K of the Carnot refrigerator will be,

K = |Q_L|/W

   = T_L/T_H-T_L

Here T_L is lower temperature, T_H is the higher temperature, W is the work done by the engine and Q_L is the heat at lower temperature.

To obtain the efficiency ε of the Carnot engine, substitute T₂ for T_L, T₁ for T_H and Q₁ for Q_H in the equation ε = 1- T_L/T_H =|W|/|Q_H|,

ε =1-T₂/T₁

  = |W|/|Q₁|

To obtain the coefficient of performance K of the Carnot refrigerator, substitute T₄ for T_L, T₃ for T_H and Q₄ for Q_L in the equation K = T_L/T_H-T_L= |Q_L|/W,

K =T₄/T₃-T₄

    = |Q₄|/|W|

Lastly, |Q₄| = |Q₃| - |W|.

We just need to combine these three expressions into one.

1-T₂/T₁ = |W|/|Q₁|

|W| = |Q₁| T₁-T₂/T₁

From K =T₄/T₃-T₄ =|Q₄|/|W|, we get,

T₄/T₃-T₄ = |Q₄|/|W|

              = |Q₃| - |W|/|W|     (Since, |Q₄| = |Q₃| - |W|)

              = |Q₃|/|W| - 1

Substitute the value of |Q₁| T₁-T₂/T₁ for W in the equation T₄/T₃-T₄ =|Q₃|/|W| - 1, we get,

T₄/T₃-T₄ =|Q₃|/|W| - 1

              =|Q₃|/|Q₁| (T₁/T₁-T₂)- 1

So, |Q₃|/|Q₁| = (T₄/T₃-T₄ +1) (T₁-T₂/T₁)

                     = (T₃/T₃-T₄) (T₁-T₂/T₁)

                     = (1- T₂/T₁) / (1- T₄/T₃)

From the above observation we conclude that, the ratio|Q₃|/|Q₁| in terms of the four temperatures would be equal to (1- T₂/T₁) / (1- T₄/T₃).

______________________________________________________________________________________

Problem:

One mole of an ideal monoatomic gas is caused to go through the cycle as shown in below figure. Hw much work is done on the gas in expanding the gas from a to c along path abc? (b) What is the change in internal energy and entropy in going from b to c? (c) What is the change in internal energy and entropy in going through one complete cycle? Express all answers in terms of the pressure p₀ and volume V₀ at a point a in the diagram.

Concept:-

Work done W at constant pressure is defined as,

W = pΔV

Here p is the pressure and ΔV is the change in volume.

Change in internal energy ΔE is defined as,

ΔE = 3/2 nRΔT

Here n is the number of moles, R is the gas constant and ΔT is the change in temperature.

Change in entropy ΔS is defined as,

ΔS = 3/2 ln (T_f/T_i)

Here, T_f is the final temperature and T_i is the initial temperature.

Solution:

(a) To obtain the work done W_abc along the path abc, first we have to find out the work done W_ab along the path ab.

From the figure we observed that,

 W_ab = pΔV

         = p₀ (V₀-4V₀)

         = -3 p₀V₀

So,   W_abc= W_ab

                = -3 p₀V₀

From the above observation we conclude that, the work done on the gas along the path abc would be -3 p₀V₀.

ΔE_bc = 3/2 nRΔT_bc

         = 3/2 (nRT_c – nRT_b)

         = 3/2(p_cV_c – p_bV_b)         (pV = nRT)

         = 3/2[(2p₀) (4V₀) -(p₀) (4V₀)]

         = 3/2 [8 p₀V₀ - 4 p₀V₀]

          = 6 p₀V₀

The change in entropy ΔS_bc along the path b to c would be,

ΔS_bc = 3/2nR ln(T_c/T_b)

        = 3/2nR ln(P_c/P_b)       (Since, T_c/T_b = P_c/P_b)

        =3/2nR ln(2P₀/P₀) 

        = 3/2 nR ln2  

From the above observation we conclude that, the change in internal energy ΔE_bc in the path b to c would be 6 p₀V₀ and the change in entropy ΔS_bc in the path b to c would be 3/2 nR ln2.