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

Answer:

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

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"#