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.
We know the specific heat of the iron is (say
Cp1) ~ 450 J/kg.
and specific heat of water is (Say Cp2) ~
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
=> m1Cp1(T1-T) = m2Cp2(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*Cp1*ln(T/T1) = 1*450*ln(306.22/1173) =
-604.36 J/K
change of entropy (say
s2) of water is = m*Cp2*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
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