Supose the material of the skillet is iron.
The specific heat constant for iron is: \(c_{Fe} = 470 \frac{J}{kg \cdot K}\)
Since the unit of measurement for the specific heat constant is Joules per kilogram Kelvin, we transform the temperatures from degrees Celsius to degrees Kelvin:
T1 = 12 °C = 273.15 + 12 = 285.15 K
T2 = 84 °C = 273.15 + 84 = 357.15 K
We also convert the mass from grams to kilograms:
m = 46 g = 0.046 kg
The amount of heat absorbed by the skillet is:
\(Q = m \cdot c_{Fe} \cdot \Delta T = m \cdot c_{Fe} \cdot (T_2 - T_1)\)
Replacing the above values, we get:
\(Q = 0.046 \cdot 470 \cdot (357.15 - 285.15) = 1557\)
The answer is 1557 J (joules) or 1.557 kJ (kilojoules).
If the material is aluminium, use the specific heat of aluminium in your calculation, which is:
\(c_{Al} = 920 \frac{J}{kg \cdot K}\)
