Question

4.5 kg of water at 23 ∘C is mixed with 4.5 kg of water at 46 ∘C in a well-insulated container.

Estimate the net change in entropy of the system

Answer 1

T = equilibrium temperature

T_h = temperature of hot water = 46

m_c = mass of cold water = m_h = mass of hot water= 4.5 kg

T_c = temperature of cold water = 23

c = specific heat = 4186

using conservation of heat

Heat lost by hot water = heat gained by cold water

m_h c (T_h - T) = m_c c (T - T_c)

(4.5) (4186) (46 - T) = (4.5) (4186) (T - 23)

T = 34.5

for cold water :

S_c = entropy change for cold water = m C_p ln(T/T_c) = (4.5) (4.2 x 10³) ln((34.5 + 273)/(23 + 273)) = 720.4

for hot water :

S_h = entropy change for hot water = m C_h ln(T/T_h) = (4.5) (4.2 x 10³) ln((34.5 + 273)/(46 + 273)) = - 694

Total entropy change is given as

S = S_c + S_h = 720.4 - 694 = 26.4