How much heat is absorbed by a 72g iron skillet when its temperature rises from 80C to 260C?

The correct answer is 5819.04 J

The energy (q) required to elevate an object of mass (m) and specific heat capacity (c) by a temperature (T) is given by q = mcΔT.

Here, m is 72 g, c is 0.449 J/Kg °C, ΔT = 260 - 80 = 180 °C

Now, putting the values we get,

q = 72 * 0.449 * 180

q = 5819.04 J of energy is absorbed by the skillet.