What is the specific heat capacity of a 17.5g sample of an unknown metal that releases 6700J of heat when it cools from 90 to 25 degrees C?

1 Answer
Jul 8, 2017

The specific heat capacity of the metal is (6color(white)(.)"J")/("g"*""^@"C")6.JgC.

Explanation:

You will use the calorimetry formula:

q=cmDeltaT,

where q is energy in Joules (J), c is the specific heat capacity, m is mass, and DeltaT is the change in temperature. DeltaT=(T_"final"-T_"initial")

Organize your data:

Known

q"="-"6700 J" lArr q is negative because energy was released

m="17.5 g"

DeltaT=(T_"final"-T_"initial")=(25^@"C"-90^@"C")=-65^@"C"

Unknown

c

Solution

Rearrange the formula above to isolate c. Insert your data into the formula and solve.

c=(q)/(m*DeltaT)

c=(-6700"J")/(17.5"g"xx-65^@"C")=(6color(white)(.)"J")/("g"*""^@"C") rounded to one sig fig because of 90^@"C"