Reversible and Irreversible Process:-

Problem 1:

Tweleve grams of Nitrogen (N₂) in a steel tank are heated from 25.0 to 125ºC. (a) How many moles of nitrogen are present? (b) How much heat is transferred to the nitrogen?

Solution:

Number of moles (n) is equal to given mass of molecule (m) divided by molecular weight (W) of the molecule,

n = m/M

The heat transferred (Q) in a constant volume process is equal to,

Q = nC_vΔT

Here C_v is the molar heat capacity at constant volume, ΔT (ΔT =T_f –T_i) is the raising temperature and n is the number of moles.

Thus, Q = nC_vΔT

              = nC_v(T_f –T_i)

n = m/M

    = (12 g)/(28 g/mol)

  = 0.429 mol

Thus 0.429 moles of nitrogen are present.

To obtain the heat Q which is transferred to the nitrogen, substitute 0.429 mol for n, 5/2 R for C_v and 125° C for T_f and 25^° C for T_i in the equation Q = nC_v(T_f –T_i),

Q = nC_v(T_f –T_i)

   = (0.429 mol) (5/2 R) (125° C-25^° C)

  = (0.429 mol) (5/2 ×8.31 J/mol. K) ((125 + 273) K-(25 + 273) K)      (Since, R = 8.31 J/mol. K)

 = (0.429 mol) (5/2 ×8.31 J/mol. K) (398 K-298 K)

 = (0.429 mol) (5/2 ×8.31 J/mol. K) (100 K)

= 891 J

From the above observation we conclude that, the heat Q which is transferred to the nitrogen would be 891 J.

_______________________________________________________________________

Problem 2:

A thermometer of mass 0.055 kg and heat capacity 46.1 J/K reads 15.0ºC. 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 ºC, what was the temperature of the water before insertion of the thermometer, neglecting other heat losses?

Solution:

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.

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

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.

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

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:

An aluminum electric kettle of mass 0.560 kg contains a 2.40 kW heating element. It is filled with 0.640 L of water at 12.0ºC. How long will it take (a) for boiling to begin and (b) for the kettle to boil dry? (Assume that the temperature of the kettle does not exceed 100ºC at any time.)

Solution:

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,

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

Q= mc (T_f - T_i)

   = mcΔT

m = ρV

Time t is equal to the heat energy Q divided by power P.

t = Q/P

t = Q/P

 = [m_ac_a +ρ_wV_wc_w]ΔT/P

 = [m_ac_a +ρ_wV_wc_w] (T_f - T_i)/P

Here, mass of aluminum electric kettle is m_a, specific heat of aluminum is c_a, density of water is ρ_w, volume of water is V_w, specific heat of aluminum is c_w, final temperature is T_f , initial temperature is T_i and power is P.

To obtain the time t taken by the electric kettle for boiling to begin, substitute 0.56 kg for m_a, 900 J/kg. K for c_a, 998 kg/m³ for ρ_w, 0.640 L for V_w, 4190 J/kg.K for c_w, 100°C for T_f and 12^° C for T_i and 2.40 kW for P in the equation t =[m_ac_a +ρ_wV_wc_w] (T_f - T_i)/P,

t =[m_ac_a +ρ_wV_wc_w] (T_f - T_i)/P

 = [(0.56 kg) (900 J/kg. K) + (998 kg/m³) (0.640 L) (4190 J/kg.K)] (100°C -12^° C)/2.40 kW

= [(0.56 kg) (900 J/kg. K) + (998 kg/m³) (0.640 L×10^-3 m³/1 L) (4190 J/kg.K)] ((100+273) K –(12+273) K)/(2.40 kW×10³ W/1 kW)

= [(0.56 kg) (900 J/kg. K) + (998 kg/m³) (0.640×10^-3m³) (4190 J/kg.K)] (373 K –285 K)/(2400 W)

= 117 s

From the above observation we conclude that, the time t taken by the electric kettle for boiling to begin would be 117 s.

(b) The time t for the kettle to boil dry will be,

t = Q/P

 = ρ_wV_wL_w/P

Here density of water is ρ_w, volume of water is V_w, latent heat of water is L_w and power is P.

To obtain the time t for the kettle to boil dry, substitute 998 kg/m³ for ρ_w, 0.640 L for V_w, 2256×10³ J/kg for L_w and 2.40 kW for P in the equation t = ρ_wV_wL_w/P,

t = ρ_wV_wL_w/P

 = (998 kg/m³) (0.640 L) (2256×10³ J/kg)/( 2.40 kW)

= (998 kg/m³) (0.640 L×10^-3 m³/1 L) (2256×10³ J/kg)/( 2.40 kW×10³ W/1 kW)

= (998 kg/m³) (0.640 ×10^-3 m³) (2256×10³ J/kg)/( 2400 W)

= 600 s

From the above observation we conclude that, the time t for the kettle to boil dry would be 600 s.

Question 1