An electrical heater is used to supply 85.0 J of energy to a 10.0 g sample of aluminum, originally at 295.0 K. Determine the final temperature in degrees Kelvin. (the specific heat capacity for aluminum is 24.35 Jmol-1K-1)
Molar mass of Al = 26.98 g/mol
mass(Al)= 10.0 g
use:
number of mol of Al,
n = mass of Al/molar mass of Al
=(10 g)/(26.98 g/mol)
= 0.3706 mol
Use:
Q = n*C*(Tf-Ti)
85.0 J = 0.3706 mol * 24.35 J/mol.K * (Tf - 295.0) K
Tf - 295.0 = 9.42
Tf = 304 K
Answer: 304 K
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