The initial temperature of a bomb calorimeter is 28.50°C. When a chemist carries out a reaction in this calorimeter, its temperature decreases to 27.45°C. If the calorimeter has a mass of 1.400 kg and a specific heat of 3.52 J/(gi°C), how much heat is absorbed by the reaction? Use q equals m C subscript p Delta T..
140 J
418 J
1,470 J
5,170 J

The correct answer is 5,170 J

Explanation:

The heat absorbed by the reaction (Qr) is equal to the heat released by the calorimeter (Qcal):

Qr = - Qcal

We calculate Qcal from the mass of the calorimeter (m), the specific heat (Cp), and the change in temperature (ΔT), as follows::

Qcal = m x Cp x ΔT

We have the following data:

m = 1.400 kg x 1000 g/1 kg = 1,400 g

Cp = 3.52 J/g°C

Initial T = 28.50°C + 273 = 301.50 K

Final T = 27.45°C + 273 = 300.45 K

ΔT = Final T - Initial T = 300.45 K - 301.50 K = -1.05 K

So, we introduce the data in the mathematical expression for Qcal:

Qcal = m x Cp x ΔT = 1,400 g x 3.52 J/g°C x (-1.05 K) = -5,174.4 J ≅ -5,170 J

Therefore, the heat absorbed by the reaction is:

Qr = - Qcal = -(-5,170 J) = 5,170 J

it's c. elements are placed in rows based on how may valencies electrons have.