ThermodynamicsHard
Question
A piece of alloy weighing 4 kg and at a temperature of 800 K is placed in 4 kg of water at 300 K. If the specific heat capacity of water is 1.0 cal/K-g and that of alloy is 4 cal/K-g, then the ΔSmix is (ln 2 = 0.7, ln 3 = 1.1, ln 7 = 1.95)
Options
A.+3.33 kcal/K
B.−1.0 kcal/K
C.+1.0 kcal/K
D.+1.33 kcal/K
Solution
Heat lost by alloy = Heat gained by water
or 4 × 4 × (800 – T) = 4 × 1.0 × (T – 300) ⇒ T = 700 K
(As date is not given for vaporization of water)
Now, $\Delta S_{mix} = \Delta S_{alloy} + \Delta S_{water}$
$= 4 \times 4 \times \ln\frac{700}{800} + 4 \times 1 \times \ln\frac{700}{300} = 1.0\text{ K cal/K}$
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