Given 132.8j of energy is required to heat 11.17g of aluminum from 15.73°c to 28.94°c find the specific heat of aluminum.

0.9 J/g.°C.

Explanation:

Q = m.c.ΔT,

where, Q is the amount of energy (Q = 132.8 J).

m is the mass of Al (m = 11.17 g).

c is the specific heat capacity of Al (c of Al = ??? J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 28.94°C - 15.73°C = 13.21°C).

∵ Q = m.c.ΔT

∴ (132.8 J) = (11.17 g)(c)(13.21°C).

∴ c = (132.8 J)/(11.17 g)(13.21°C) = 0.9 J/g.°C.

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