Answer: Step 1: Calculate qsur (the surrounding is
usually the water)
qsur = ? J
m = 75.0 g water
c = 4.184 J/g
oC
ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC
qsur = m · c · (ΔT)
qsur = (75.0g) (4.184 J/g
oC) (-9.4 oC)
qsur = - 2949.72 J
First, using the information we know that we
must solve for qsur, which is the water. We know
the mass for water, 75.0g, the specific heat of
the water, 4.184 j/g
o
c, and the change in
temperature, 21.6-31.0 = -9.4 oC. Plugging it
into the equation, we solve for qsur.
Step 2: Calculate qsys qsys = - (qsur)
qsys = - (- 2949.72 J)
qsys = + 2949.72
In this case, the qsur is negative, which means
that the water lost energy. Where did it go? It
went to the system. Thus, the energy of the
system is negative, opposite, the energy of the
surrounding.
Step 3: Calculate moles of the substance
that is the system
Given: 12.8 g KCl
Mol system = (g system given)
(molar mass of system)
Mol system = (12.8 g KCl)
(39.10g + 35.45g)
Mol system = 12.8 g KCl
74.55 g
Mol system = 0.172
Here, we solve for the mol in the system by
using the molar mass of the material in the
system.
Step 4: Calculate ΔH ΔH = q sys .
Mol system
ΔH= + 2949.72 J
0.172 mol
ΔH= +17179.81 J/mol or +1.72 x 104
J/mol
i hope this helps
= 1.1 kg/L
= 1100 g/L