Question

In an experiment, 150 g of aluminum (with a specific heat of 900 J/kg·K) at 67.0°C is mixed with 69.0 g of water (with a specific heat of 4186 J/kg·K) at 18.0°C, with the mixture thermally isolated. (a) What is the equilibrium temperature? What are the entropy changes of (b) the aluminum, (c) the water, and (d) the aluminum-water system?

Answer 1

Mass of aluminium m_Al = 0.15 Kg , C_Al = 900 J/kg·K T₁ = 67.0°C

Mass of water m_w= 0.069 Kg , Cw = 4186 J/kg·K T₂ = 18.0°C

Let T be the equilibrium temperature

Heat lost = Heat Gained

m_Al x  C_Al x (T1 - T) = m_w x Cw x (T - T2)

0.15 x 900 x ( 67 - T) = 0.069 x 4186 x ( T - 18)

Solving T = 33.60 C (Ans a)

Entropy change of aluminium = m_Al x  C_Al x log_e ((T+273) / (T1+273)) = -13.95 J/K Ans b

Entropy change of water = mw x  Cw x log_e ((T+273) / (T2+273)) = 13.95 J/K = 15 J/K Ans c

Entropy change of Al- water system =  Entropy change of aluminium + Entropy change of water = 1.05 J/K Ans d