91Ó°ÊÓ

Specific heat (c) of the iron =0.451 J/g °C.

Weight of the iron skillet = 5.00×10 ²-g = 500g.

From the specific heat of the iron, it is clear that 0.451 g of heat is required to heat 1g of iron by 1°C.

By using the equation below, we can calculate heat in joules.

\(\begin{array}{c}{\bf{q = c \times m \times \Delta T}}\\{\bf{ = c \times m \times (}}{{\bf{T}}_{{\bf{final}}}}{\bf{ - }}{{\bf{T}}_{{\bf{initial}}}}{\bf{)}}\end{array}\)

The known values (Step 1) are substituted as shown below.

\(\begin{array}{c}{\rm{q = c \times m \times (}}{{\rm{T}}_{{\rm{final}}}}{\rm{ - }}{{\rm{T}}_{{\rm{initial}}}}{\rm{)}}\\\begin{array}{*{20}{l}}{{\rm{ = 0}}{\rm{.451 J/g ^\circ C \times 500g \times }}\left( {{\rm{250 ^\circ C - \;25^\circ C}}} \right)}\\{{\rm{ = 0}}{\rm{.451 J\; \times 500 \times }}\left( {{\rm{225}}} \right)}\\\begin{array}{l}{\rm{ = \;0}}{\rm{.451 \times 112500 }}\\{\rm{ = 50, 737}}{\rm{.5 J }}\\{\rm{ = 5}}{\rm{.05 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{J}}\end{array}\end{array}\end{array}\)

Thus, the required amount of heat, in Joules, is \({\rm{5}}{\rm{.05 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{J}}\).