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Physical Property Notes Specific Heat (Cp) Inquiry: As a small group, discuss and answer the following questions: 1) Do different substances heat up at different rates? Give me an example to support your answer: 2) Do different substances cool down at different rates? Give me an example to support your answer: 3) Why do you think this happens? Specific heat capacity (Cp) = the amount of heat/energy necessary to change the temperature of 1.00 gram of a substance by 1.00 °C Some things heat up and cool down fast…like sand on the beach in the summer time These substances have ____________ specific heat Some things heat up and cool down slowly…like water in the ocean in the summer time These substances have ____________ specific heat Calculations q = mCp∆T q=heat gained or lost ∆T = temperature change of substance *Always for 1 gram of substance m = mass of substance *Always only 1oC change in temp Cp = specific heat capacity of substance *Substances have known specific heats that are constant – look them up in your reference table! *Units: J/g-oC ***Super Important Concept How does heat/energy flow? High to Low or Low to High ???? If a hot metal is placed into cold water, describe how heat/energy will flow So…if the hot metal loses 10 J of energy does the water gain 10 J? _______ ***Super Important Concept ______________________________ How do we calculate specific heat in the lab? 1) Heat up a known mass of a metal in boiling water until the substance has reached the same temperature as the boiling water (100oC) 2) Transfer (very quickly) the hot metal to a water bath that contains a specific amount of water at a known temperature (usually the temperature of the room) 3) Measure the temperature change in the water bath **We must make the assumption that all heat/energy lost by the hot metal is transferred and gained by the water** qlost by metal = qgained by water mmetalCpmetal∆Tmetal = mH2OCpH2O∆TH2O 4) Solve for unknown Cp of the metal using known Cp of water and other given data Example 1: 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. Example 2: A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 °C. Assuming no loss of energy to the surroundings: 1. How many joules of energy did the water absorb? 2. How many joules of energy did the metal lose? 3. What is the specific heat of metal?