how much heat is required to raise the temperature of 475 grams of solid aluminum from room temperature (295 K) to liquid aluminum at 1050K? aluminum melts at 933.47 K

Answer 1

The total amount of heat needed: ∑q = q₁ + q₂ + q₃

q₁ = mCΔT = 475 g x 0.902 J/g-^oC x (933.47 - 295) K = 273,552.47J = 273 kJ (in solid stage)

ΔT = (933.47 - 295) K = 638.47 K ΔT = (933.47 K -273) - (295 K - 273) = 638.47 oC

^{For ΔT temperature in K is same as in oC}

q₂ = mol ΔH_f (in melting stage)

mol = 475 g x 1 mol /27g = 17.6 mol q2 = mol ΔHf = 17.6 mol x 10.79 kJ/mol = 189.9 kJ

q3 = mCΔT = 0.475 kg x 1.18 kJ/kg-oC (1050 - 933.47)K = 65.3 kJ

∑q = q₁ + q₂ + q₃ = 273 + 189.9 + 65.3 = 528.2 kJ

Answer 2

You need to know the specific heat for aluminum, which is 0.902 J/g/deg. You also need to know the ∆H_fusion, which is ~ 400 J/g.

1. heat to change 475 g Al from 295K to 933.47K = q = mC∆T = (475g)(0.902 J/g/deg)(638.47deg) = 273,552 J
2. heat to melt 475 g Al at 1050K = q = (m)(∆H_fusion) = (475 g)(400 J/g) = 190,000 J
3. heat to raise temp of 475 g Al from 933 to 1050K= q = mC∆T = (475g)(1.18 J/g/deg)(117K)= 65,579J
4. Total heat = 273,552J + 190,000 J + 65,579 J = 529,131 J = 529 kJ (to 3 significant figures)