Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K)
© 2006 Brooks/Cole - Thomson
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Change in Temperature (T)
∆T of System Increase Sign of ∆Tsystem + Tfinal >Tinitial Tfinal < Tinitial Sign of q + Direction of Heat Transfer Heat transferred from surroundings to system (_____thermic) Heat transferred from system to surroundings (_____thermic)
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Decrease
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© 2006 Brooks/Cole - Thomson
Chapter 6 — Thermo — Part 1
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13 14 Example 1: If 25.0 g of Al cool from 310oC to 37oC, how many joules of heat energy are lost by the Al?
Heat Transfer q = m x Cs x ∆T q = C x ∆T
• • • • • q = heat absorbed or released (J) Cs = specific heat (J g-1 K-1) C = heat capacity (J K-1) m = mass of substance (g) ∆T = change in temperature (K) = Tfinal - …show more content…
H2O(g) + C(graphite) → H2(g) + CO(g) Honet = To convert 1 mol of water to 1 mol each of H2 and CO requires _______of energy. The “water gas” reaction is ______thermic.
© 2006 Brooks/Cole - Thomson
In general, when ALL enthalpies of formation are known,
Horxn = Σ Hfo(products) - Σ Hfo(reactants)
Remember that
© 2006 Brooks/Cole - Thomson
always = final – initial
Using Standard Enthalpy Values
Calculate the heat of combustion of methanol, CH3OH(g) + 3/2 O2(g) → CO2(g) + 2 H2O(g) Horxn = Σ Hfo (product) - Σ Hfo (reactant) Horxn = Hfo (CO2) + 2 Hfo (H2O) - {3/2 Hfo (O2) + Hfo (CH3OH)}
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BORN-HABER Cycle
• It is an energy cycle • Lattice energies cannot be measured directly, but can be found by using the BORN-HABER cycle • Lattice energy can be obtained indirectly by applying Hess’s Law in Born-Haber cycle
= (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} Ho rxn = -675.6 kJ per mol of