Solved Problems on Thermodynamics:-

Problem 1:-

A container holds a mixture of three nonreacting gases: n₁ moles of the first gas with molar specific heat at constant volume C₁, and so on. Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantitites of the three separate gases. 

Concept:-

Heat capacity C of a body as the ratio of the amount of heat energy Q transferred to a body in any process to its corresponding temperature change ΔT.

C = Q/ΔT

So, Q = C ΔT

Each species will experience the equal temperature change.

If the gas has n molecules, then Q will be,

Q = nC ΔT

Solution:-

If the gas has n₁ moles, then the amount of heat energy Q₁ transferred to a body having heat capacity C₁ will be,

Q₁ = n₁C₁ ΔT

Similarly, if the gas has n₂ moles, then the amount of heat energy Q₂ transferred to a body having heat capacity C₂ will be,

Q₂ = n₂C₂ ΔT

And

if the gas has n₃ moles, then the amount of heat energy Q₃ transferred to a body having heat capacity C₃ will be,

Q₃ = n₃C₃ ΔT

As, each species will experience the same temperature change, thus,

Q = Q₁ + Q₂ + Q₃

   = n₁C₁ ΔT + n₂C₂ ΔT + n₃C₃ ΔT

Dividing both the sides by n = n₁ + n₂ + n₃ and ΔT, then we will get,

Q/nΔT = (n₁C₁ ΔT + n₂C₂ ΔT + n₃C₃ ΔT)/ nΔT

As, Q/nΔT = C, thus,

          C=ΔT (n₁C₁ + n₂C₂  + n₃C₃)/ nΔT

          = (n₁C₁ + n₂C₂  + n₃C₃)/ n

          = n₁C₁ + n₂C₂  + n₃C₃/ n₁ + n₂ + n₃

From the above observation we conclude that, the molar specific heat at constant volume of the mixture would be n₁C₁ + n₂C₂  + n₃C₃/ n₁ + n₂ + n_3.

Problem 2:-

A thermometer of mass 0.055 kg and heat capacity 46.1 J/K reads 15.0°C. It is then completely immersed in 0.300 kg of water and it comes to the same final temperature as the water. If the thermometer reads 44.4°C, what was the temperature of the water berfore insertion of the thermometer, neglecting other heat losses?

Concept:-

In accordance to the law of conservation of energy, for a thermodynamic system, in which internal is the only type of energy the system may have, the law of conservation of energy  may be expressed as,

Q + W = ΔE_int

Here Q is the energy transferred between the system and its environment, W is the work done on or by the system and ΔE_int is the change in the internal energy of the system.

The heat capacity C of a body as the ratio of amount of heat energy Q transferred to a body in any process to its corresponding temperature change ΔT.

C = Q/ΔT

So, Q = C ΔT

The heat capacity per unit mass of a body, called specific heat capacity or usually just specific heat, is characteristic of the material of which the body is composed.

c = C/m

  = Q/mΔT

So, Q = c mΔT

Solution:-

The heat transfers for the water Q_w is,

Q_w = m_wc_w (T_f –T_i)

Here, mass of water is m_w, specific heat capacity of water is c_w, final temperature is T_f and initial temperature is T_i.

The heat transfers for the thermometer Q_t is

Q_t = C_tΔT_t

Here, heat capacity of thermometer is C_t and ΔT_t is the temperature difference.

As the internal energy of the system is zero and there is no work is done, therefore substitute ΔE_int = 0 and W = 0 in the equation Q + W = ΔE_int,

Q + W = ΔE_int

 Q + 0= 0

So, Q = 0

Or, Q_w + Q_t = 0

m_wc_w (T_f –T_i)+ C_tΔT_t = 0

So, T_i = (m_wc_w T_f + C_tΔT_t )/ m_wc_w

Here ΔT_t = 44.4 ^° C - 15.0 ^° C

               =  29.4 ^° C

To obtain the temperature of the water before insertion T_i of the thermometer, substitute 0.3 kg for m_w, 4190 J/kg.m for c_w, 44.4 ^° C for T_f, 46.1 J/K for C_t and 29.4 ^° C for ΔT_t in the equation T_i = (m_wc_w T_f + C_tΔT_t )/ m_wc_w,

T_i = (m_wc_w T_f + C_tΔT_t )/ m_wc_w

   = [(0.3 kg) (4190 J/kg.m) (44.4 ^° C) + (46.1 J/K) (29.4 ^° C)] /[(0.3 kg) (4190 J/kg.m)]

  =45.5 ^° C

From the above observation we conclude that, the temperature of the water before insertion of the thermometer was 45.5 ^° C.

Problem 3:-

A mixture of 1.78 kg of water and 262 g of ice at 0°C is, in  a reversible process, brought to a final equilibrium state where the water / ice ratio, by mass 1:1 at  0°C. (a) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics. 

Concept:-

The entropy change ΔS for a reversible isothermal process is defined as,

ΔS = Q/T

     = -mL/T

Here m is the mass, L is the latent heat and T is the temperature.

Solution:-

(a) Mass of water = 1.78 kg

Mass of ice = 262 g

So the total mass of ice and water mixture will be,

Mass of ice-water mixture = (Mass of water) + (Mass of ice)

                                            = (1.78 kg) + (262 g)

                                            = (1.78 kg) + (262 g×10^-3 kg/1 g)

                                            = 1.78 kg + 0.262 kg

                                            = 2.04 kg

If eventually the ice and water have the same mass, then the final state will have 1.02 kg (2.04 kg/2) of each.

Thus the mass of the water that changed into ice m will be the difference of mass of water m_w  and mass of final state m_s.

So, m = m_w - m_s

To obtain mass of water that changed into ice m, substitute 1.78 kg for mass of water m_w and 1.02 kg for mass of final state m_s in the equation m = m_w - m_s,

m = m_w - m_s

   = 1.78 kg – 1.02 kg

  = 0.76 kg

The change of water at 0^° C to ice at 0^° C  is isothermal.

To obtain the change in entropy ΔS of the system during this process, substitute 0.76 kg for mass m, 333×10³ J/kg for heat of fusion of water L and 273 K for T in the equation ΔS = -mL/T,

ΔS = -mL/T

     = -(0.76 kg) (333×10³ J/kg )/(273 K)

     = -927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system during this process will be -927 J/K.

(b) Now the system is returned to the first equilibrium state, but in an irreversible way. Thus the change in entropy ΔS of the system during this process is equal to the negative of previous case.

So, ΔS = -(- 927 J/K)

           = 927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system would be 927 J/K.

(c) In accordance to second law of thermodynamics, entropy change ΔS is always      zero.

The total change in entropy will be,

ΔS = (-927 J/K) + (927 J/K)

      = 0

From the above observation we conclude that, our answer is consistent with the second law of thermodynamics.  

Problem 4:-

Apparatus that liquefies helium is in a laboratory at 296 K. The helium in the apparatus is at 4.0 K. If 150 mJ of heat is transferred from the helium, find the minimum amount of heat delivered to the laboratory.

Concept:-

Coefficient of performance K of a Carnot refrigerator is defined as,

K = T_L / T_H - T_L        …… (1)

Here T_L is the lower temperature of sink and T_H is the higher temperature of source.

K = (what you want)/(what you pay for)

   = Q_L/W

and this is called coefficient of performance. The larger the value of K, the more efficient is the refrigerator.

Thus, W = Q_L/K      …… (2)

Substitute the value of K from equation (1) in the equation W = Q_L/K,

W = Q_L/K

    = Q_L/( T_L / T_H - T_L)

    = Q_L (T_H/ T_L – 1)       …… (3)

W = Q_H – Q_L

Here Q_H is the exhausted heat.

Thus exhausted heat will be,

Q_H = W + Q_L     …… (4)

Substitute the value of W from equation (3) in the equation Q_H = W + Q_L,

Q_H = W + Q_L    

      = Q_L (T_H/ T_L – 1) + Q_L

     = Q_L (T_H/ T_L)       

To obtain the minimum amount of heat delivered to the laboratory, substitute 150 mJ for Q_L, 296 K for T_H and 4.0 K for T_L in the equation Q_H = Q_L (T_H/ T_L),

Q_H = Q_L (T_H/ T_L)

      = ((150 mJ) (10^-3 J/1 mJ)) (296 K/4.0 K)

      = 11 J

From the above observation we conclude that, the minimum amount of heat delivered to the laboratory would be 11 J.

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