A 40kg piece of steel (Cp = .5kJ kg^-1 K^-1) at temperature 450C is quenched in

Question

A 40kg piece of steel (Cp = .5kJ kg^-1 K^-1) at temperature 450C is quenched in 150kg of oil (Cp = 2.5kJ kg^-1 K^-1) at temperature 25C. If there are no heat loses what iss the change of entropy of (a) the steel, (b) the oil. and (c) both together. Hint: you need to find the new temperature after the steel and iol have reached thermal equilibrium. You can do tat by assuming that the total energy (which energy is that?) does not change, while the energy change for each material is due only to the change in temperature.

Explanation / Answer

let the common temperature be T

therefore, Q= m* Cp * delta T

or 40 * 0.5 * ( 450 - T) = 150 * 2.5 * ( T-25 ) ...

or 20 * 450 - 20 T = 375 T - 375 * 25

or T = 46.52 C = 319.51 K

now, a) entropy = m*Cp * ln ( T2/T1) = 40 * 0.5 ln(319.51/723) = -16.332 kJ/K

b) entropy =m*Cp * ln ( T2/T1) = 150*2.5 ln(319.51/298 ) = 26.136 kJ/K

c) total entropy = -16.332 + 26.136 = 9.804 kJ/K

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