Question

In a calorimetry experiment, a student forgot to weigh a sample of lead that she used in the experiment. She heated the lead to 350.^oC and immediately placed the heated sample into a polystyrene calorimeter containing 475. gm H₂O at 22.5^oC. The final temperature was 35.7^oC. What was the mass of the lead sample?

Specific heat capacity of lead is: 0.121 J/gm/^oC

Specific heat capacity of water: 4.184 J/gm/^oC.

Answer 1

Sol :-

Amount of heat gain by water = Mass of water x specific heat capacity of water x change in temperature

= 475 g x 4.184 J/g/^oC x (35.7 - 22.5) ^oC

= 26233.68 J

Because,

Amount of heat release by lead = Amount of heat gain by water

So,

26233.68 J = Mass of lead x specific heat capacity of lead x change in temperature

26233.68 J = Mass of lead x 0.121 J/g/^oC x (35.7 - 350) ^oC

Mass of lead = 26233.68 J / 0.121 J/g/^oC x (35.7 - 350) ^oC

= 689.8 g

= 690 g (approx.)

Hence, mass of lead sample was = 690 g