1. Q1: A lead pellet having a mass of 26.47 g at 89.98ºC was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50 ºC to 23.17 ºC. What is the specific heat for the lead pellet? [Answer: 280.3 J]
Calorimeter is an isolated system (no heat lost to the surrounding).
Therefore, we can write: qPb + qH2O = 0 OR qPb = -qH2O
The heat gained by water is given by
qH2O = msΔT = (100 g) (4.184 J/g.ºC) (23.17 ºC – 22.50 ºC)
= 280.3 J
= 280.3 J
Because the heat lost by lead pellet is equal to the heat gained by water, so qPb = - 280.3 J
qPb = msΔT thus, -280.3 J = (26.47 g) (s) (23.17 ºC – 89.98 ºC)
s = 0.158 J/g.ºC
2. Q2: A manufacturer claims that its new dietetic dessert has fewer than 10 Calories per serving. To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2. If the heat capacity of the calorimeter is 8.151KJ/K and the temperature increases 4.937 ºC, is the manufacturer’s claim correct?
[Answer: the claim is correct]
-q sample = q calorimeter
q calorimeter = msΔT = (8.151 kJ/K) (4.937 K) = 40.24 kJ.
Recall that 1 Calorie = 1 kcal = 4.184 kJ. Therefore, 10 Calories = 41.84 kJ. So, the claim is correct!!