A 100.0g sample of lead is heated to 100.0 o C (Celsius) and is placed in a coff

Question

A 100.0g sample of lead is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 25.0 oC. After the metal cools, the final temperature of the metal and the water is 26.4 oC. Calculate the specific heat capacity of lead from these experimental data, assuming that no heat escapes to the surroundings or is transferred to the calorimeter.
Specific heat of water = 4.184 J/g oC
atomic mass Pb = 207.2 g/mol
Calculate your answer in J/g oC with 3 significant figures, but enter the answer without units.

Then calculate the molar heat capacity of Pb in J / mol oC, again, to 3 significant figures. Enter the answer without units.

Explanation / Answer

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

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