A sample of 5.23 kg of aluminum absorbs heat,which raises the temperature 50.0°C.Determine the amount of heat absorbed,in kJ,if the specific heat capacity of aluminum is 0.903 J/g°C.

Question

Quizplus · Accepted Answer

First, we need to convert the mass of aluminum from kg to g:

5.23 kg x 1000 g/kg = 5230 g

Then, we can use the formula:

q = m x c x ∆T

where q is the amount of heat absorbed, m is the mass of the substance, c is the specific heat capacity, and ∆T is the change in temperature.

Plugging in the values we know:

q = 5230 g x 0.903 J/g°C x (50.0°C) = 236,053 J

To convert J to kJ, we divide by 1000:

q = 236.053 kJ

Therefore, the answer is C) 236 kJ.

A Sample of 5