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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?