| Basic Thermodynamics ~ J. Pohl © | www.THERMOspokenhere.com (175-D225) |
The Indians of the early southwest used stone boiling to cook tender parts of bison, particularly the liver. After a successful hunt the paunch and liver were cut from a bison. Fire wood was gathered and stacked in criss-crossed layers then stones were placed on top and the wood was set afire. Nearby, the paunch was inverted, filled with water and hung on a tripod of branches.
 |
As one Indian sliced liver, others replenished the fire until the stones became very hot. The cooking set-up with the initial conditions of the stones, liver and water are sketched. Finally using sticks the "cook" lifted the hot stones from the fire and dropped them, along with sliced liver, into the bag of water. Determine the second temperature of the rock, water and liver in the bag.
♦ We take the rock, water and liver (approximated as an equal mass of water) as system. Ignore the thermal effects of the paunch. A precise second temperature cannot be determined because there will be heat to the surroundings. The best we can do at this level is to approximate the event as adiabatic and calculate the greatest possible second temperature of the liver and water. There will be negligible kinetic and potential energy changes. The energy equation and its reduction for this event is:
| (1) 1 |
The rock is initially hot (500°C) and the water and liver are cold (10°C). After they are together, a second, uniform temperature will be attained. We expect the final temperature to be in the range: 10°C < T2 < 500°C. The rock and liver will remain solid and liquid respectively. We don't know if the water will boil. To proceed, an assumption is required.
Assume the final phase of water is liquid.
Water and liver are assumed to have the same properties so we combine them.
| (2)2 |
We will use specific heats to solve. The first step is to construct the enthalpy differences. Since we have assumed no phase change the enthalpies of the second states will equal those of the initial state plus a change. These are written as follows:
| (3)3 |
Enter these differences into the energy equation:
| (4) 4 |
While we clean this equation we keep all terms "left of equality." There is no heat. The work is "atmosphere compresses" - this is a constant pressure thermal equilibration event. Take the above equation a step further, the initial enthalpies sum to zero leaving only the energy changes - stone relative to water (as written below).
| (5) 5 |
The second temperature will surely be less than 500°C. The first term, the energy change of the rock, is negative, as it should be - the energy of the rock decreases. Also the "water plus liver" term is positive. We can put numbers in this now.
| (6)6 |
From this we determine the second temperature to be, T2 = 258°C! But this is impossible since it is well above 100oC, the boiling temperature of water at 1 atmosphere. We conclude our assumption that the water remained liquid was wrong.
Assume the final phase of water is liquid and vapor.
To be careful, we will treat the water and liver separately. We write a mass equation for the water which expresses that liquid and gas exist in the final state.
| (7)7 |
Next we "postulate" the energy changes.
| (8)8 |
By our assumption of liquid and vapor water in the second state, the second temperature will be 100°C which we use in the next equation. We have kept the terms in position. Next we enter data for specific heats and the enthalpy boiling, hfg
| (9) 9 |
The only unknown quantity becomes the mass of the water (mg) that is vapor upon conclusion of the assumed adiabatic event. By careful calculation, the above equation yields the amount of gas (or vapor) created as:
mg = 2.4 kg which seems reasonable. Consequently the second temperature is T2 = 100°C.
