Final answer:
To heat 244 g of aluminum from 22.2 °C to 39.3 °C, one would require 3.76 kJ of heat energy, using the specific heat capacity of aluminum and the mass and temperature change.
Step-by-step explanation:
To calculate the amount of heat required to heat 244 g of Al(s) from 22.2 °C to 39.3 °C, we can use the specific heat capacity of aluminum and the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The specific heat capacity (c) for aluminum, according to given data, is 900 J/kg°C, which we need in terms of g°C, so we convert it to 0.900 J/g°C.
The mass (m) is already given as 244 g. The change in temperature (ΔT) is 39.3 °C - 22.2 °C = 17.1 °C.
Now, plugging in the values we get Q = 244 g * 0.900 J/g°C * 17.1 °C
= 3755.64 J.
Since the question asks for the answer in kJ, we convert the Joules to kJ by dividing by 1000. Thus, the final answer is 3.76 kJ.