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Whitepunk [10]
4 years ago
13

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 h

eat capacity of the lead is what?
Chemistry
1 answer:
Kryger [21]4 years ago
3 0

Answer:

0.13 J/g.°C.

Explanation:

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

<em>Q = m.c.ΔT,</em>

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

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

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