CALCULATING SPECIFIC HEAT
CAPACITY
Using lab data and your
understanding of
thermal energy!
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.
The Key Equation

qlost by the metal = qgained by the water
So…
mmetalx Cmetal x∆tmetal = mwaterx Cwaterx∆twater.
(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.

Now you try one…


An 800 g chunk of lead is heated in boiling
water @100 oC, for 5 min. It is placed into
250 g of cold water @ 12.5 oC. Within 2
min., the water and lead reach thermal
equilibrium at 20.0 oC.
Calculate the specific heat capacity of lead
based on this data.
Here’s another…


Suppose 100 g of aluminum were heated in
99.0 oC water, then quickly transferred into a
foam cup holding 107 g of 10.0 oC water.
The final temperature recorded is 24.0 oC.
What is the specific heat capacity of
aluminum based on this data?
Check your answers!



clead = .122 J/g -oC
caluminum = .836 J/g –oC
NOTE: It is helpful to think of specific heat
capacity as resistance to change in
temperature – sort of like a thermal inertia!