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?
2.25 J/ (°C × g).
1.75 J/ (°C × g).
0.444 J/ (°C × g).
0.324 J/ (°C × g).

Answer 1

Heat gained by the metal = heat lost by the water
65.0* Cs * (100.0°C - 27.0°C) = 100.8 g * 4.18 J/(°Cxg) * (27.0°C - 22.0°C)

Cs = 100.8 * 4.18 * 5.0 / (65.0 * 73) J/(°Cxg) = 0.444J/(°Cxg)

Answer: the fhird option, 0.444 J/(°Cxg)