Question

In a foundry, metal castings are cooled by quenching in an oil bath. Typically, a casting weighting 20 kg and at a temperature of 450 oC is cooled by placing it in a 150 kg in volatile oil bath initially at 50 oC. If the specific heat capacity of the metal is 0.5 J/kg K, and that of the oil is 2.6 J/kg K, determine the entropy change in this process and the common final temperature of the oil and the casting after quenching. Assume that there are no heat losses

Answer 1

4.18 J/KgK

Step-by-step explanation:

Equilibrium point is reached when

m₁*c₁(T₁-T) = m₂*c₂(T -t₂)

m1 = 20

c1 = 0.5

T1 = 450

m2 = 150 kg

c2 = 2.6

T2 = 50

putting these values into the formula

20*0.5 (450-T) = 150*2.6(T - 50)

4500 - 10T = 390T - 19500

4500 + 19500 = 390T + 10T

24,000 = 400T

T = 24000/400

= 60⁰C

ΔSmetal = m1*c1In[t + 273]/[T1+273]

= 20*0.5 In (60+273)/450+273

= 10 ln(333/723)

= 10 * -0.7752

= -7.752

ΔS/oil =

m2*s2(60 + 273)/50 + 273)

= 150*2.6ln(333/323)

= 390 * 0.03048

= 11.88j/KgK

Δtotal = -7.7+11.8

= 4.18J/KgK

this is the enthropy change