Answer:
72·45 °C
Explanation:
Given
Mass of sample of aluminum is 500·0 g
Initial temperature is 25·0 °C = 298 K
Heat absorbed from surroundings is 32·60 kJ
Heat capacity of aluminum is 0·9 J/g K
Formula for amount of heat transfer is
<h3>Q = m × s × ΔT</h3>
where
Q is the amount of heat transfer
m is the mass of the substance
s is the heat capacity of the substance
ΔT is the change in temperature
Let the final temperature be T K
∴ 32·60 × 1000 = 500 × 0·9 × (T - 298)
T - 298 = 72·45
∴ T = 370·45 K = (370·45 - 298) °C = 72·45 °C
∴ Final temperature is 72·45 °C
The overall enthalpy of reaction will be calculated from enthalpy of formation of products and reactants.
DeltaHrxn = [Sum of enthalpy of formation of products] - [sum of enthalpy of formation of reactants]
DeltaHrxn = 3 X deltaHf (H2O) - [ 3 X DeltaH (H2) + DeltaH (O3)]
DeltaHrxn = 3 X (--242) - [ 0 + 143] = -869 kJ / mole
We can also calculate the enthalpy of reaction from bond energies
The first step is to convert the 55 kg of Fe₃O₄ to moles by using the molar mass, that is, 231.55 g/mol (55 kg = 55000 gms),
55000 gms × 1 mole / 231.55 gms = 237.53 moles of Fe₃O₄
For the decomposition of every 1 mole of Fe₃O₄, +1118 kJ of energy is needed. Therefore, the energy needed to decompose 237.53 moles will be:
237.53 moles of Fe₃O₄ × 1118 kJ / 1 mole = 265558 kJ
Thus, there is a need of 265558 kJ of energy to decompose 55 kg of Fe₃O₄.
Concentration is wrong, conc=moles/volume, both concentration are wrong use the formula to correct it.