Question

The temperature of a 15-g sample of lead metal increases from 22 °C to 37 °C upon the addition of 29.0 J of heat. The specific heat capacity of the lead is what?

Answer 1

Answer:

0.13 J/g.°C.

Explanation:

• To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat absorbed by lead (Q = 29.0 J).

m is the mass of lead (m = 15.0 g).

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

ΔT is the temperature difference (final T - initial T) (ΔT =  37 °C - 22 °C = 15.0 °C).

∵ Q = m.c.ΔT

∴ c = Q/m.ΔT = (29.0 J)/(15.0 g)(15.0 °C) = 01288 J/g.°C ≅ 0.13 J/g.°C.