Worked example 1
How much heat is needed to raise 125 g of water from 21.0 C to 68.0 C? (c = 4.18 J/g C)
Try it first: Write DT before touching the calculator.
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Thermochemistry & ThermodynamicsCalorimetryContent level: Core 24 min
Specific Heat and Calorimetry Calculations
What you'll be able to do: Use q = mcDT confidently, including problems where two substances exchange heat and reach a common final temperature.
Best after: Energy, Heat and the First Law
Introduction
One equation carries most of this section. The difficulty is never the algebra, it is knowing which mass, which specific heat and which temperature change belong in each term.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Apply q = mcDT to a single substance
- Interpret specific heat capacity and compare substances
- Solve heat-exchange problems using q(hot) + q(cold) = 0
- Track units and signs through a multi-step calorimetry calculation
Lesson
The central equation
Heat depends on how much substance there is, what it is made of, and how far its temperature moved. Those three factors are the mass, the specific heat capacity and the temperature change.
q = m c DT, DT = T(final) - T(initial)
What specific heat means
Specific heat is the energy in joules needed to raise one gram of a substance by one °C. Water is 4.18 J/g C, aluminium about 0.897, copper about 0.385. A low specific heat means the substance heats and cools quickly, which is why a metal spoon burns your hand long before the soup does.
A temperature change in Celsius is numerically identical to one in kelvin, so specific heats in J/g C and J/g K are interchangeable.
Two objects, one final temperature
When a hot metal is dropped into cool water, energy leaves the metal and enters the water until both sit at the same temperature. The heat lost equals the heat gained, so the two q values sum to zero once signs are included.
m(metal) c(metal) (Tf - Ti,metal) + m(water) c(water) (Tf - Ti,water) = 0
Both terms use the same final temperature Tf but different initial temperatures. Writing DT as Tf - Ti automatically gives the metal term a negative value.
Choosing the mass
For a reaction in solution, the mass in q = mcDT is the mass of the solution, not the mass of the solute alone. Dilute aqueous solutions are normally assumed to have the density and specific heat of water unless the problem says otherwise.
Unit discipline
Specific heats are quoted per gram, so masses must be in grams. Answers come out in joules; divide by 1000 for kilojoules. Enthalpies of reaction are almost always reported in kJ/mol, which needs a mole calculation as a separate final step.
Key ideas
Definition
Specific heat capacity
Energy required to raise the temperature of 1 g of substance by 1 C, in J/g C.
Definition
Heat capacity of a calorimeter
Energy needed to raise the whole apparatus by 1 C, in J/C, with no mass term.
Rule
DT direction
Always compute final minus initial so the sign of q comes out automatically.
Rule
Heat exchange
For an isolated pair of objects, q(hot) + q(cold) = 0.
Key concept
Why water is used
Its high specific heat means it absorbs a lot of energy for a small, easily measured temperature rise.
Equation
Heat and temperature change
q = m c DT
- m = mass in grams
- c = specific heat capacity in J/g C
- DT = T(final) - T(initial) in C
Equation
Calorimeter constant form
q = C(cal) DT
- C(cal) = heat capacity of the entire calorimeter in J/C
Worked examples
Worked example 2
A 55.0 g piece of copper at 99.5 C (c = 0.385 J/g C) is dropped into 100.0 g of water at 22.0 C. Find the final temperature.
Try it first: Set the two heat terms to sum to zero and keep one unknown Tf.
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Worked example 3
A calorimeter with heat capacity 18.5 J/C rises by 4.20 C during a reaction. How much heat did the calorimeter itself absorb?
0 of 2 steps revealed.
Common mistakes
Writing DT as initial minus final.
Why it's wrong: It reverses the sign of every q in the problem.
Check instead: Always use final minus initial.
Using kilograms with a specific heat in J/g C.
Why it's wrong: The answer comes out 1000 times too large.
Check instead: Convert mass to grams first.
Using only the solute mass for a reaction in solution.
Why it's wrong: The whole solution is what changes temperature.
Check instead: Use the total solution mass, taking 1.00 g/mL if a density is not given.
Expecting the final temperature to be the average of the two starting temperatures.
Why it's wrong: That is only true when the two masses and specific heats match.
Check instead: Weight the result toward the substance with the larger mc product.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
The heat absorbed or released by a substance is q = mcDT, where DT is always final minus initial. Specific heat is the energy needed to raise one gram by one degree; water is unusually high at 4.18 J/g C, which is why it makes a good calorimeter fluid. When a hot object is dropped into cooler water, the heat lost by one equals the heat gained by the other, so q(hot) + q(cold) = 0, and the common final temperature falls between the two starting temperatures.
- q = mcDT with DT taken as final minus initial
- Specific heat is per gram, so masses belong in grams
- Water at 4.18 J/g C dominates most calorimetry calculations
- For heat exchange, the two q values sum to zero
- Use the mass of the whole solution, not just the solute
Sources and further reading
- Chemistry 2e, Section 5.1: Energy BasicsPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 5.1View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License
- Chemistry 2e, Section 5.2: CalorimetryPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 5.2View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License