Answer to Question #177631 in Chemistry for Kelsy

Question

What is the specific heat of aluminum if the temperature of an 85.2g sample of aluminum is increased by 24.3º, when 621J of heat is added?

Answer 1

Solution:

q = m × C × ΔT

where:

q = amount of heat energy gained or lost by substance (J)

m = mass of sample (g)

C = specific heat capacity (J ^oC^-1 g^-1)

ΔT = change in temperature (^oC)

 

Thus:

(621 J) = (85.2 g) × C × (24.3º)

С = (621 J) / (85.2 g × 24.3^oC) = 0.2999 J ^oC^-1 g^-1 = 0.300 J ^oC^-1 g^-1

C(Al) = 0.300 J ^oC^-1 g^-1

NB! Although the specific heat of aluminum should be 0.900 J ^oC^-1 g^-1 (table value).

Answer: The specific heat of aluminum is 0.300 J ^oC^-1 g^-1