A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water rises to 27.0° C. If the specific heat capacity of liquid water is 4.18 J / (°C * g), what is the specific heat of the metal?

Question

Chemistry

Lamont Richard

28 September, 21:45

A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water rises to 27.0° C. If the specific heat capacity of liquid water is 4.18 J / (°C * g), what is the specific heat of the metal?

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Kameron Gordon · Answer 1

We do a heat balance to solve this:

(m cp ΔT) water = - (m cp ΔT) metal

100.8 (4.18) (27 - 22) = - 65 (cp) (27-100)

cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))

cp = 0.44 J / (°C * g)

The specific heat of the metal is 0.44 J / (°C * g)