Question

How much heat is absorbed by a 46g skillet when its temperature rises from 12C to 84C

Answer 1

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:
$c_{Al} = 920 \frac{J}{kg \cdot K}$