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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CalorimetryCourse level: High School / HonorsCourse level: AP Chemistry
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
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.
Use ΔT = Tfinal − Tinitial with its sign. Cooling 50.0 g of water from 35.0 to 20.0 °C gives q = (50.0)(4.18)(−15.0) = −3135 J: the water releases about 3.14 kJ. A temperature interval of 15.0 °C equals 15.0 K, so either interval unit is acceptable if it matches c. Absolute kelvin conversion is needed for thermodynamic T in Gibbs or gas equations, not for this subtraction.
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, for a given mass and heat input; how quickly heat travels through a spoon also depends strongly on thermal conductivity.
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.
Write each temperature change using the shared unknown Tf: m₁c₁(Tf − T₁) + m₂c₂(Tf − T₂) = 0. Rearrangement gives Tf = (m₁c₁T₁ + m₂c₂T₂)/(m₁c₁ + m₂c₂). The weights are heat capacities mc, not masses alone. The final temperature lies between the starting temperatures and closer to that of the body with larger heat capacity. Include Ccal(Tf − Tcal,initial) if the cup's heat capacity is significant. Heat loss to the room would require an additional term.
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.
| Supplied property | Units | Heat expression |
|---|---|---|
| Specific heat c | J/(g·K) | q = mcΔT, with m in g |
| Molar heat capacity Cm | J/(mol·K) | q = nCmΔT |
| Whole-object heat capacity C | J/K | q = CΔT |
Read the denominator before multiplying by mass. If c varies strongly with temperature, a constant-c calculation is an approximation. During a phase change at its transition temperature, heat can be absorbed without a temperature rise: q = nΔHfus or nΔHvap. A warming–melting–warming path needs separate sensible-heat and phase-change terms. Do not use mcΔT across a melting plateau as though the entire input raised temperature.
Use heat capacity to compare temperature changes
Specific heat c is the energy required per gram per degree. At a fixed heat input and mass, a material with smaller c undergoes a larger temperature change.
Check whether the problem supplies a specific heat in J/(g·°C) or an entire calorimeter heat capacity in J/°C; only the former is multiplied by mass.
q = mcΔT for a material; q = CcalΔT for a calibrated calorimeter.
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.
Worked example 4
An 80.0 g metal sample with c = 0.400 J/(g·°C) absorbs 640 J. How much does its temperature rise?
Try it first: Solve q = mcΔT for ΔT.
0 of 3 steps revealed.
Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Writing DT as initial minus final. | It reverses the sign of every q in the problem. | Always use final minus initial. |
| Using kilograms with a specific heat in J/g C. | Using the numerical mass in kilograms with c in J/(g·°C) makes q 1000 times too small. | Convert mass to grams first. |
| Using only the solute mass for a reaction in solution. | The whole solution is what changes temperature. | 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. | That is only true when the two masses and specific heats match. | Weight the result toward the substance with the larger mc product. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
34 topic questions across 3 difficulty levels.
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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
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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
Lesson-specific rights and provenance need review.
The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.
Recorded use: reference. This describes the stored citation, not verified authorship or reuse permission.
Recorded attribution / license claim (applicability unverified): Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0.
Source rights and review status - Chemistry 2e, Section 5.2: CalorimetryPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 5.2View source
Lesson-specific rights and provenance need review.
The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.
Recorded use: reference. This describes the stored citation, not verified authorship or reuse permission.
Recorded attribution / license claim (applicability unverified): Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0.
Source rights and review status