WS: Calorimetry Problems

1. How much energy is needed to change the temperature of 50 1. How much energy is needed to change the temperature of 50.0 g of water 15.0 oC? Q = m x s x T Q = (50.0g)(4.18 J/g oC)(15.0 oC) = 3135 J

2. How many grams of water can be heated from 20.0 oC to 75oC using 12.5 kilojoules? m = Q/s x T 12.5 kJ (1000 J/1kJ) = 12,500 J T = (75oC - 20oC) = 55oC m = (12,500 J)/(4.18 J/g oC)(55.0 oC) = 54.4 g

3. What is the final temperature after 840 Joules is absorbed by 10 3. What is the final temperature after 840 Joules is absorbed by 10.0 g of water at 25.0 oC? T = Q/m x s T = (840 J)/(10.0 g)(4.18 J/g oC) = 20.1 ºC 25 ºC + 20.1 ºC = 45.1 ºC

4a. How much energy is needed to raise the temperature of a 8 4a. How much energy is needed to raise the temperature of a 8.50 x 102g block of aluminum from 22.8oC to 94.6oC? (Al = 0.90 J/g oC) Q = m x s x T 8.50 x 102 g = 850 g T = (94.6 – 22.8) = 71.8 oC Q = (850 g)(0.90 J/g oC)(71.8 oC) = 54927 J or 54.9 kJ

4b. What is the specific heat capacity of aluminum per mole? Al = (0.90 J/g oC)(27.0 g/mol) = 24.3 J/mol oC)

5. A piece of metal weighing 59. 047 g was heated to 100 5. A piece of metal weighing 59.047 g was heated to 100.0 °C and then put it into 100.0 mL of water (initially at 23.7 °C). The metal and water were allowed to come to an equilibrium temperature, determined to be 27.8 °C. Assuming no heat lost to the environment, calculate the specific heat of the metal.

5. Determine heat gained by water: Q = m x s x T Density of water = 1.0 g/ mL Q = (100.0g)(4.18 J/g oC)(4.1 oC) = 1714 J Since no heat loss Qwater = Qmetal

5. Determine specific heat of metal: S = Q/m x T T = 100 ºC – 27.8 ºC = 72.2 ºC S = 1714 J/(59.047 g)(72.2 ºC) = 0.402 J/g ºC