11 Oct 2011


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

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!!                      

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