1. Lab 12: Heat, Energy, and Temperature This is it!! Today we are going to measure the specific heat of an unknown metal. Important terms: Temperature, T: a measure of a body’s hotness or coldness. An measure of the average kinetic energy of the molecules Heat, Q: a from of energy used to do work. Specifically, heat is the energy that must be transferred Specific Heat, c: the amount of heat required to raise the the temperature of an object by one degree Celsius.  Q = change in heat  T = change in temperature m = mass of the object How much heat is required to raise 5 kg of water for 20 degrees C to 80 degrees C?

2. This equation is not valid when a substance undergoes a “change of phase” (ice  water, water  vapor) When heat energy is added to change to phase of a substance this energy is used to break the bonds of the molecules. During this process, the temperature does not change. Consider a 1 kg block of ice at –50 C. Slowly adding heat causes the ice to warm. Let’s say we warm the ice until it reaches 0 C. Energy Temperature -50 C 0 C ice

3. Once the temperature reaches ) C, the temperature will not change as more heat energy is added. All the will happen is that the ice will change phase. The heat required to melt the ice is called the latent heat of fusion, f. So, the 333000 J is the energy needed to change the ice completely into water. Energy Temperature -50 C 0 C ice ice + water mix Once we have water at 0 C, we will continue to add heat energy to heat the water to 100 C

4. Energy Temperature -50 C 0 C ice ice + water mix 100 C water water + steam  Q= m V

5. Today we are going to do two experiments. Power = Energy/time = E/t so then E = Power x time = P t Here T i is the initial temperature. You are essentially going to measure the temperature rise of the water as a function of the heating time. Temperature Time the slope in this region is equal to:

6. The next experiment will allow you to measure the specific heat of a metal. You will take a hot piece of metal and drop it into a cool cup of water, from the temperature increase you can work out the specific heat. Consider a piece of metal, an aluminum cup containing some mass of water. The metal starts out with an initial temperature equal to, while the cup and water start out at the same temperature, BEFOREAFTER When the metal is dropped into the water, all three objects equalize to the same temperature, T f TfTf

7. Since this is a closed system, the water and cup gain the energy lost by the meta. So that means:l Solving for the specific heat of the metal gives us: