Chapter4, Lesson C - Quenching a Copper Ingot in Oil

4C-3 :	Quenching a Copper Ingot in Oil							3 pts

A copper ingot of volume 1 L is heat treated at 500°C and then quenched in a 100 L oil bath initially at 20°C. Assuming no heat exchange with the surroundings, determine the temperature of the ingot and the oil when they reach thermal equilibrium.

For copper:	ρ = 8900 kg/m3,		CV = 0.386 kJ/kg-K
For oil:	ρ = 910 kg/m3,		CV = 1.8 kJ/kg-K

Read :	The easiest way to solve this problem is to choose the entire contents of the tank, both the oil and the copper, as our system. If we assume that this system is adiabatic and does not have any work interactions with its surroundings, then the internal energy of the system must remain constant as the copper ingot cools and the oil becomes warmer. If we further assume that the copper and oil are incompressible, then this is a constant volume process. For solids and liquids it is often reasonable to assume the heat capacity is a constant over a fairly wide temperature range. The only unknown left in the 1st Law is the final system temperature !







Given:	VCu =	0.001	m3			ρCu =	8900	kg/m3
	TCu,1 =	500	°C			CV,Cu =	0.386	kJ/kg-K
	Voil =	0.1	m3			ρoil =	910	kg/m3
	Toil,1 =	20	°C			CV,oil =	1.8	kJ/kg-K

Find:	T2 =	???	°C

Assumptions:		Copper and oil have constant heat capacities. No heat is exchanged with the surroundings by either the copper or the oil. Copper and the oil are both incompressible, so this process is a constant volume process.

Solution :	We begin by writing the 1st Law and we choose as our system the oil and the copper.

								Eqn 1

	By cleverly selecting our system, Q = 0 and W = 0. This makes the solution simpler.

		Eqn 2		Therefore:	mCu =	8.9	kg
					moil =	91	kg

	Because both oil and copper are assumed to be incompressible with constant heat capacities:							Eqn 3

								Eqn 4

	Now, solve for T2:

								Eqn 5

								Eqn 6

	Now, plug values into Eqn 6 to evaluate T2 :

Answer:	T2 =	29.9	oC

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