SME
Grade 10 Science · Problem type
Calorimetry: q = mcΔT
Use this for “A reaction warms 100 g of water from 21 °C to 27.5 °C. How much heat did it give off?”, or a metal’s specific heat from its heat, mass and warming.
Diagram
Heat taken in by the water: q = mcΔT = 100 × 4.18 × 6.5 = 2,720 J
The water warmed: the reaction gave off 2,720 J.
Exothermic.
This is an example. Type your own values to start a new problem.
q (J)
=
m (g)
×
c (J/(g·°C))
×
(
T₂ (°C)
−
T₁ (°C)
)
ΔT
Change in temperature
calculated · °C
Formulas
ΔT
=
T₂
−
T₁
6.5 = 27.5 − 21
q
=
m
×
c
×
ΔT
2,717 = 100 × 4.18 × 6.5
Assumptions
•
The foam cups keep heat from getting in or out, so the water takes in all of it.
•
Water’s specific heat is 4.18 J/(g·°C).
•
The heat the water takes in is the heat the reaction gives off.
Step-by-step
WE KNOW
m = 100 g, c = 4.18 J/(g·°C), T₁ = 21 °C, T₂ = 27.5 °C
FIND
change in temperature (ΔT), heat taken in (q)
Step 1 · Find change in temperature (ΔT)
Use ΔT = T₂ − T₁
The change is the final temperature minus the first.
ΔT = 27.5 − 21
ΔT = 6.5 °C
Step 2 · Find heat taken in (q)
Use q = m × c × ΔT
Heat is mass times specific heat times the change in temperature.
q
=
100
×
4.18
×
6.5
The reaction gave off 2,717 J (qᵣₓₙ = −2,717 J)
q
=
2,717
J
Check
Put the numbers back in. Both sides should match.
6.5 = 27.5 − 21 ✓
2,717 = 100 × 4.18 × 6.5 ✓
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Thermochemistry: heat, enthalpy and calorimetry
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