# "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?

$124 \text{J/kg·K}$
Heat energy required is given by $W = m c \Delta T$.
therefore c= W/(mDeltaT)=(1351"J")/((0,125"kg")(112,1^"o""C"-25.0^"o""C"))="124 J/kg·K"