SME
Grade 10 Science · Problem type
A cold pack: ΔH from a calorimeter
Use this for the ΔH per mole of a salt that cools the water as it dissolves.
Diagram
Heat taken in by the water: q = mcΔT = 100 × 4.18 × −6.1 = −2,550 J
The water cooled: the reaction took in 2,549.8 J from it.
Endothermic.
This is an example. Type your own values to start a new problem.
m
Mass of water
example · g
c
Specific heat of water
example · J/(g·°C)
T₁
Starting temperature
example · °C
T₂
Final temperature
example · °C
ΔT
Change in temperature
calculated · °C
q
Heat the water takes in
calculated · J
qᵣₓₙ
Heat of the reaction
calculated · J
n
Moles of salt
example · mol
ΔH
Enthalpy change per mole
calculated · kJ/mol
Formulas
ΔT
=
T₂
−
T₁
−6.1 = 15.9 − 22
q
=
m
×
c
×
ΔT
−2,550 = 100 × 4.18 × (−6.1)
qᵣₓₙ
=
−
q
2,550 = −(−2,550)
ΔH
=
(
qᵣₓₙ
1,000
)/
n
25.5
=
(
2,550
1,000
)/0.1
Assumptions
•
The solution’s heat capacity is taken as the water’s, 4.18 J/(g·°C).
•
The heat the reaction takes in is the heat the water loses: qᵣₓₙ = −q.
•
ΔH is per mole of salt, in kJ: divide by 1000.
Step-by-step
WE KNOW
m = 100 g, c = 4.18 J/(g·°C), T₁ = 22 °C, T₂ = 15.9 °C, n = 0.1 mol
FIND
change in temperature (ΔT), heat the water takes in (q), heat of the reaction (qᵣₓₙ), enthalpy change per mole (ΔH)
Step 1 · Find change in temperature (ΔT)
Use ΔT = T₂ − T₁
The change is the final temperature minus the first.
ΔT = 15.9 − 22
ΔT = −6.1 °C
Step 2 · Find heat the water takes in (q)
Use q = m × c × ΔT
Heat is mass times specific heat times the change in temperature.
q
=
100
×
4.18
×
(−6.1)
q
=
418
×
(−6.1)
q
=
−2,550
J
Step 3 · Find heat of the reaction (qᵣₓₙ)
Use qᵣₓₙ = −q
What the water loses, the reaction takes in.
qᵣₓₙ = −(−2,550)
qᵣₓₙ = 2,550 J
Step 4 · Find enthalpy change per mole (ΔH)
Use ΔH = (qᵣₓₙ ÷ 1,000)/n
Change the joules to kJ (1 kJ = 1000 J), then share the heat among the moles.
ΔH
=
(
2,550
1,000
)/0.1
ΔH = 2.55/0.1
ΔH = 25.5 kJ/mol
Check
Put the numbers back in. Both sides should match.
−6.1 = 15.9 − 22 ✓
−2,550 ≈ 100 × 4.18 × (−6.1) ✓
2,550 = −(−2,550) ✓
25.5
=
(
2,550
1,000
)/0.1
✓
Topic check
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Thermochemistry: heat, enthalpy and calorimetry
Main lesson
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