Friday, January 6th

Today started off as a normal day in Mr.H's 3rd period chemistry class. We went over Tim's blog that was on enthalpy that we had done the previous day. Next Mr.H directed us to page 4 in our packets where we found an ordinary problem about determining the heat released in the combustion of a walnut, determining the heat content of a walnut and determining the amount of calories in 50 walnuts.

This problem started with the heat released by the combustion of a walnut. We began with the equation q=mCΔT where q=the amount of heat released, m=the mass of the heated substance C=the specific heat(4.18 for water) and T for the temperature change. In this problem we needed to find the amount of heat that was released into the water causing the water to change temperature. We began plugging in the known values in the equation and ended up with a multiplication problem of:q=(200g)(4.18)(39.1-25). Then after the multiplying ended up with 11787.6J. This was the amount of heat gained by the water. Since the water gained energy and the walnut lost it this makes the heat lost by the walnut -11787.6J. this can be expressed in the equation: qwalnut=-qH2O, meaning as the walnut looses energy the walnut gains it.

The next part of the problem was to find the energy content of the walnut in kJ/gram. for this we set up a conversion factor that converted from J to kJ then we divided by the amount of grams in the problem to determine the kJ/gram. This conversion factor looked like:(11787.6J)(1kJ/1000J)= 11.7876kJ the diving by 2.6 making the answer 4.53 kJ/ gram.

The Final part of the problem was the amount of calories in 50 walnuts. This problem consisted of a few simple conversions from kJ to Calorie. First, we know that 11787.6 J are released in one walnut so then in 50 walnuts there will be 50 times as much heat:(50 walnuts)(11787.6/1 walnut)=58930J of heat. Then we must convert this back to kJ because it as given in the problem that 4.18 kJ=1 calorie:(58930J)(1kJ/1000J)= 58.930kJ of heat. Last we set up the conversion factor to finish the problem:(58.930kJ)(1 cal/4.18kJ)= ~141calories in 50 walnuts.

Next we moved on to the next problem on page 7 of our packets problem number 1 which consisted of STOICHIOMETRY. YAYYYY. SWEEEETTTTT

This problem was the combustion of iron oxide.

4Fe(s)+3O2(g) -> 2 Fe2O3(s) ΔH=-1652kJ

a. For part a, we needed to find the energy released by the combustion of 1 mol Fe. This was a simple conversion factor that converted 1 mole Fe to the amount of energy that would be theoretically produced. this conversion factor looked like (1 mole Fe)(1652kJ/4 mole). this resulted in a answer of 413 for the first problem.

b. Next for part b, this was almost the same thing as part a. but with a different amount of Fe. the conversion looked like (12 mole Fe)(1652kJ/4 mole). this parts answer was 4956 kJ.

c. Last part c. this one was a little harder because of the conversion of grams to moles, but still was simple. this conversion looked like (110g Fe)(1 mole/55.845g Fe)(1652kJ/4 mole)= 813.5 kJ as the final answer.

This was the end of the lesson part of the class. Next, we were on the the lab test part of class. This lab test was to combust paraffin in the for of candle. Due to the Acedemic dishonesty policy at GBS. I may not divulge more information than that. (i dont want to cheat or help someone cheat). The lab took us to the end of the period.

Overall it was a good day. =P