|4C-3 :||Quenching a Steel Bar in Oil||4 pts|
|A hot steel bar weighing 20 kg is submerged in an insulated bath holding 50 L of heavy oil. The steel bar and the oil are allowed to equilibrate thermally without exchanging heat with the surroundings.|
|Initially, the steel bar and the oil are at 700oC and 25oC, respectively. Determine the final temperature of the steel bar and the oil.|
|Data: steel: r = 8000 kg/m3, CV = 0.42 kJ/kg-K, oil: r = 890 kg/m3, CV = 2.1 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 steel, 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 steel bar cools and the oil becomes warmer.
If we further assume that the steel 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 !
|Diagram:||The diagram in the problem statement is adequate.|
|Assumptions:||1 -||- Steel and oil have constant heat capacities.|
|2 -||- No heat is exchanged with the surroundings by either the steel or the oil.|
|3 -||- Steel and the oil are both incompressible, so this process is a constant volume process.|
|Equations / Data / Solve:|
|We begin by writing the 1st Law and we choose as our system the oil and the steel.|
|By cleverly selecting our system, Q = 0 and W = 0. This makes the solution simpler.|
|Because both oil and steel are assumed to be incompressible with constant heat capacities:|
|Now, solve for T2:||
|Verify:||None of the assumptions can be verified from the data given in the problem statement.|
|Assumptions 1 & 3 are very nearly true for solids over the temperature range covered in this problem.|
|Assumption 2 could be made nearly true with sufficient insulation.|