1. Solving Specific Heat Capacity Problems 3 main types Finding Heat lost or gained (Q= mcΔT ) when there is warming or cooling Finding the heat needed to vaporize or released during condensation (Q=mL v ) Finding the heat needed to melt or released when freezing ( Q=mL f )

2. Q = mcΔT Q is thermal energy in Joules m is mass in g C is the specific heat capacity of a substance i.e. how much heat is needed to change 1 g of the substance by 1° C ΔT is the temperature change

3. Example 1 Water has a specific heat capacity of 4.18 J/gC°. How much heat is needed to raise the temperature of 100. g from 20° C to 50.°C ? Q = mcΔT so Q = (100.g)(4.18 J/gC° )(30°C) Q = 12 540 J

4. Example 2 How much heat energy is needed to change 40. g of water to steam? Q=mL v = (40. g)(2260 j/g) =90 400 J

5. Example 3 How much heat is released when 70. g of ice melts (i.e. at 0° C) Q=mL f = (70.g)(333 J/g) = 23 310 J

6. Example 4 500 J of heat energy is needed to raise the temperature of 100 g of an unknown material from 30°C to 80°C. What is the specific heat capacity of the unknown? Q = mcΔT so

7. Example 4 Solution Q = mcΔT 500. J = (100 g)(c)(80° - 30°C) 500. J = (100 g)(c)(50°C) 500. J = (5 000 gC°)(c) 500.J/(5 000 gC°) = c 0.100 J/ gC° = c

8. Example 5 A 100. g metal weight whose specific heat capacity is 0.900 J/ gC° and whose initial temperature is 95°C is placed in 50 g of water with an unknown initial temperature. If the final temperature of the water and metal is 45°C, determine the initial temperature of the water. Assume no losses of heat energy. Water’s specific heat capacity is 4.18 J/gC°. Q lost = Q gained

9. Example 5 Solution Q lost = Q gained mcΔT lost = mcΔT gained (100g)(0.900 J/ gC°)(95-45) = (50g)(4.18 J/ gC°)(45-x) 4500 J = 209 J/C°(45-x) 21.53 = 45-x, so x = 45-21.53 = 23.46°