Question

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?

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)