Blog Post #3 --- Thoughtful question

Here is my question: If the heat from burning 7.800 g of C6H6 is added to 2.387 kg of water that is initially at a temperature of 21 degrees C, what is the final temperature of the water? The specific heat of liquid is 4.184 J/g*Degrees C.

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Data:                                                                                Find:  Tf: ? of the water
   Benzene (C6H6) : 7.800 g
   H2O : 2.387 Kg
   Ti : 21.0 *C
   Specific heat of water : 4.184 J/g . C


1) First we have to find the amount of heat released by the burning of  7.800 g of benzene. To find the it, we need to first write down the balanced equation for the combustion of benzene. Then use the ∆hf* of the reactants and products  to find the ΔH°_{rxn using Hess' Law:}

ΔH°_rxn = Σ ΔH°_f (products) - Σ ΔH°_f (reactants)

  2 C6H6(l) + 15 O2(g) ----> 12CO2(g) + 6H2O(g)      ΔH°_rxn = - 6271 K

ΔH°_rxn =  12(CO₂) + 6( H₂O) – 2(C₆H₆) + 15(O₂)

ΔH°_rxn = 12(-393.5 KJ/mol) + 6 ( -241.8 KJ/ mol) – 2(49.1 KJ/ mol) + 15 (0 KJ/ mol) = -6271 KJ

2) Once we have the ΔH°_{rxn  for the balanced equation of the combustion process, we can find the amount of heat released by the burning of 7.800 g of C6H6 by simply converting from grams to moles of benzene and calculating the heat produced by the burning of that many moles of reactant.} 

7.800 g C₆H₆ x (1 mol C₆H₆ / 78.108 g) = 0.09986173 mol C₆H₆

(-6271 KJ / 2 mol C₆H₆ ) x 0.09986173 mol C₆H₆ = - 313.1165 KJ 

3) Now that we know the amount of heat produced by the  combustion of 7.800g of benzene, we are ready to plug in the values we were given and the heat we just found to solve for the ∆T  of the water using the formula q = m x C_s x ∆T; where q is the amount of heat in J, m is the mass of the substance (water) in g, C_{s is the specific heat capacity, and} ∆T is the change in temperature in °C (Nivaldo J. Tro 241). Notice that q is positive because the heat is being absorbed by the water, therefore is an endothermic process. 

q = m x C_s x ∆T

313116.5 J = (2387 g) x (4.184 J/ g . °C) x ∆T

313116.5 J/ (2387 g) x (4.184 J/ g . °C)  = ∆T

 ∆T = 31.35 °C

∆T = T_f - T_i 

∆T + T_i = T_f

T_f =  31.35 °C + 21.0 °C = 52.35 °C 

• http://www.mychemistrytutor.com/questions/calculating-delta-hrxn-combustion-benzene-using-enthalpy-formation-values
• http://www.saplinglearning.com/ibiscms/mod/ibis/view.php?id=492792
• https://sites.google.com/site/chempendix/thermo
• http://www.chemteam.info/Thermochem/HessLawIntro2.html
• Tro, Nivaldo J. “Chemistry A Molecular Approach.”Pearson. 2^nd Ed. pp. 612-655.Print.2008.