Question

A 1.00-kg iron horseshoe is taken from a forge at 900°C and dropped into 4.00 kg of water at 10.0°C. Assuming that no energy is lost by heat to the surroundings, determine the total entropy change of the horseshoe-pluswater system.

Answer 1

We know the specific heat of the iron is (say C_p1) ~ 450 J/kg. and specific heat of water is (Say C_p2) ~ 4200 J/kg.

According to the question, the mass of iron horseshoe is (say m₁) 1 kg and the mass of water is (say m₂) 4 kg.

The temperature of the iron horseshoe is (say T₁) 900 = (900+273)K = 1173K and the temperature of water is (say T₂) 10 = (10+273)K = 283K

Now, when the hot iron is dropped into the cold water, let's say the final equilibrium will be TK.

According to the conservation of heat energy (no energy is lost by heat to the surroundings),

Heat energy rejected by hot iron = heat energy absorbed by cold water

=> m₁C_p1(T₁-T) = m₂C_p2(T-T₂)

=> 1*450*(1173-T) = 4*4200*(T-283)

=> 527850 - 450T = 16800T - 4754400

=> 17250T = 5282250

=> T = 306.22

Final equilibrium temperature of the system is 306.22K

Now, change of entropy (say s₁) of iron horseshoe is = m*C_p1*ln(T/T₁) = 1*450*ln(306.22/1173) = -604.36 J/K

change of entropy (say s₂) of water is = m*C_p2*ln(T/T₂) 4*4200*ln(306.22/283) = 1324.8 J/K

Total entropy change of the horshoe-pluswater system S = s₁ + s₂ = -604.36 +1324.8 = 720.44 J/K