Calculate the heat in kJ required to convert 2.1 g of water at 52.1°C to steam at 100°C. The specific heat of water is 4.184 J/g°C and the heat of vaporization of water is 2260 J/g.

5.167 kJ

Explanation:

We have to divide the heating process into two steps: one for the heating process of liquid water (1) and the other for the phase transition from liquid water to steam at 100°C (2)

1 - heating from 52.1°C to 100°C:

heat(1) = m x Cp x ΔT = 2.1 g x 4.184 J/g°C x (100°C-52.1°C) = 420.9 J

2 - vaporization at 100°C:

heat(2) = m x ΔHv = 2.1 g x 2260 J/g = 4746 J

Finally, we add the heat values of the steps:

heat required = heat(1) + heat(2) = 420.9 J + 4746 J = 5166.9 J

Since 1 kJ= 1000 J, we convert from J to kJ:

5166.9 J x 1 kJ/1000 J = 5.1669 kJ ≅ 5.167 kJ

the answer is barium and copper bromide