When 14 grams of a sample cools from 59�C
to 27.8�C it loses 20.9 Joules of heat. What is
the specific heat of the sample?
q = mass x specific heat x (Tfinal-Tinitial)
Substitute and solve for sp. h.
to 27.8�C it loses 20.9 Joules of heat. What is
the specific heat of the sample?
Substitute and solve for sp. h.