Answer to Question #264243 in General Chemistry for Benjamin

Question

Determining the specific heat of lead :​

a- A 150.0-g sample of lead is heated to the temperature of boiling water (100.0°C).​

b- A 50.0-g sample of water is added to a thermally isolated beaker, and its temperature is found to be 22.0°C.​

c- The hot lead is dumped into cold water, and the final lead-water mixture has a temperature of 28.8°C.

Answer 1

Let c be the specific heat of Pb

Heat lost by Pb = mass x specific heat x temperature change of Pb

= 100.0 x c x (100.0 - 26.4) = 7360c J

Heat gained by water = mass x specific heat x temperature change of water

= 150 x 4.184 x (26.4 - 25.0) = 878.64 J

Total heat lost = total heat gained

7360c = 878.64

Specific heat of capacity of Pb = c = 878.64/7360

= 0.119 J/g C

Molar heat capacity of Pb = specific heat capacity x molar mass of Pb

= 0.119 x 207.2

= 24.7 J/mol C