1. CALCULATING SPECIFIC HEAT CAPACITY Using lab data and your understanding of thermal energy!

2. The Big Ideas! When in contact with each other, objects at different temperatures transfer thermal energy until they reach the same temperature. This is called thermal equilibrium. Conservation of energy requires that the thermal energy lost by the hotter object as it cools be equal to the thermal energy gained by the cooler object as it warms.

3. The Key Equation q lost by the metal = q gained by the water So… m metal x C metal x∆t metal = m water x C water x∆t water. (We should also be using some minus signs to correct for temperature decreases, but this works!) Substitute the values you know and solve for the specific heat capacity of the metal.

4. Now you try one… An 800 g chunk of lead is heated in boiling water @100 o C, for 5 min. It is placed into 250 g of cold water @ 12.5 o C. Within 2 min., the water and lead reach thermal equilibrium at 20.0 o C. Calculate the specific heat capacity of lead based on this data.

5. Here’s another… Suppose 100 g of aluminum were heated in 99.0 o C water, then quickly transferred into a foam cup holding 107 g of 10.0 o C water. The final temperature recorded is 24.0 o C. What is the specific heat capacity of aluminum based on this data?

6. Check your answers! c lead =.122 J/g - o C c aluminum =.836 J/g – o C NOTE: It is helpful to think of specific heat capacity as resistance to change in temperature – sort of like a thermal inertia!