Question

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?

Answer 1

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"`