Question

`"1351 J"` of energy are required to heat a `"0.125 kg"` sample of a metal from `25.0^@"C"` to `112.1^@"C"`. What is the specific heat of the metal?

Answer 1

`124"J/kg·K"`

Explanation:

Heat energy required is given by `W=mcDeltaT`.

`therefore c= W/(mDeltaT)=(1351"J")/((0,125"kg")(112,1^"o""C"-25.0^"o""C"))="124 J/kg·K"`