How to Calculate Heat: Q = mcΔT, Thermal Equilibrium and Latent Heat

This guide is for students learning heat in middle or high school science, anyone estimating heating or mixing in a lab report, and anyone who wants to verify a calculator result by hand.

“I cannot tell whether to use Q = mcΔT, Q = CΔT or Q = mL.” “The units get mixed up: g and kg, J and cal, K and °C.” “I never remember the mixing temperature formula.” This guide sorts out the three heat equations with worked examples. The datasets are teaching examples, not measured results.

Choose the equation by situation

Whether the temperature changes, the phase changes, or several objects are mixed determines the equation.

Q is heat (J), m is mass, c is specific heat, ΔT is the temperature change, C is heat capacity and L is latent heat. The temperature does not change during a phase change, so problems that cross a melting or boiling point are split into Q = mcΔT and Q = mL steps.

Get the units right first

There is more than one calorie. The thermochemical calorie is 1 cal = 4.184 J and the International Table calorie is 4.1868 J (NIST conversion factors). Textbooks often use 4.18 J and introductory courses 4.2 J, so use the value given in the problem and do not mix definitions. A temperature difference is the same in K and °C, but equations that need absolute temperature (such as the ideal gas law) require adding 273.15.

Example 1: warming 100 g of water by 20 K

Take the specific heat of water as 4.18 J/(g·K).

1. m = 100 g, c = 4.18 J/(g·K), ΔT = 20 K
2. Q = mcΔT = 100 × 4.18 × 20 = 8,360 J
3. 8,360 J = 8.36 kJ = 2,000 cal (using 1 cal = 4.18 J)

With the textbook value 4.2 J/(g·K), Q = 100 × 4.2 × 20 = 8,400 J. In SI units, m = 0.100 kg, c = 4186 J/(kg·K) and ΔT = 20 K give 8,372 J — the same within the spread of specific-heat values.

Specific heats in J/(g·K), converted from the OpenStax table:

Example 2: finding a specific heat

Adding 450 J to 100 g of iron raises its temperature by 10 K. Solving Q = mcΔT for c gives c = Q / (mΔT).

c = 450 ÷ (100 × 10) = 0.45 J/(g·K)

That is about one ninth of water (4.18 J/(g·K)), which is why metals heat up so much faster.

Example 3: using heat capacity

An object with heat capacity C = mc = 840 J/K warmed by 20 K needs Q = CΔT = 840 × 20 = 16,800 J. Using heat capacity avoids separating mass and specific heat. For 100 g of water, C is about 418 J/K (4.18 × 100).

Example 4: mixing two samples of water

Mix 100 g of water at 80 °C with 200 g at 20 °C. With no heat loss, the heat released by the hot sample equals the heat absorbed by the cold one.

m₁c₁(t − t₁) + m₂c₂(t − t₂) = 0, so t = (m₁c₁t₁ + m₂c₂t₂) / (m₁c₁ + m₂c₂)

1. Object A: m c = 100 × 4.18 = 418 J/K, t₁ = 80 °C
2. Object B: m c = 200 × 4.18 = 836 J/K, t₂ = 20 °C
3. t = (418 × 80 + 836 × 20) ÷ (418 + 836) = 50,160 ÷ 1,254 = 40.0 °C

The larger cold mass pulls the result towards it, so the answer is 40 °C rather than the simple average of 50 °C. Only when the heat capacities m × c are equal is it a simple average.

The same equation works for different materials. Mixing 50 g of aluminium (c = 0.90) at 100 °C with 100 g of water (c = 4.18) at 20 °C gives t = (45 × 100 + 418 × 20) ÷ (45 + 418) = 12,860 ÷ 463 ≈ 27.78 °C. The final temperature must lie between the two starting temperatures; if it does not, check the input or the assumptions.

Example 5: melting ice and boiling water (latent heat)

When 100 g of ice at 0 °C becomes water at 0 °C, the temperature does not change, so use Q = mL. With the latent heat of fusion of water at 334 J/g, Q = 100 × 334 = 33,400 J. Vaporising the same 100 g at 100 °C takes Q = 100 × 2256 = 225,600 J — about 6.8 times more heat.

When a temperature change and a phase change occur in sequence, add the steps. To turn 100 g of ice at −10 °C into water at 0 °C:

1. Warm the ice to 0 °C: Q₁ = 100 × 2.09 × 10 = 2,090 J
2. Melt it: Q₂ = 100 × 334 = 33,400 J
3. Total: Q = 2,090 + 33,400 = 35,490 J

Common mistakes

• Mixing g and kg: putting a mass in kg into J/(g·K) makes the answer 1000 times off. Match g with J/(g·K) and kg with J/(kg·K)
• Mixing calorie definitions: do not combine 4.18, 4.184 and 4.2 in one calculation
• Using Q = mcΔT during a phase change: the temperature stays constant while melting or boiling, so use Q = mL
• Misreading the sign of Q: Q > 0 means heat absorbed, Q < 0 means heat released
• Treating the mixing temperature as a plain average: when masses or materials differ, use the heat-capacity weighted average
• Comparing with a real experiment without accounting for losses: the equations assume an insulated system; real containers and air absorb heat

Check your work with the calculator

The Tools Hub heat calculator mirrors the examples above across three modes.

Applicability and limits

• The calculations assume an insulated system with no phase change unless Q = mL is used. In a real experiment heat escapes to the container and the air, so the required heat is larger than calculated
• Specific heat varies slightly with temperature; 4186 J/(kg·K) for water is a value near 15 °C
• If the process crosses 0 °C or 100 °C, split it into temperature-change and latent-heat steps and add them
• The latent heat of water also depends on pressure; use the value given in the problem
• In a mixing problem the final temperature must lie between the two starting temperatures. If it does not, the input or the assumptions (phase change, reaction, heat loss) are wrong

FAQ

What is the difference between specific heat and heat capacity?

Specific heat is the energy needed to raise 1 g of a material by 1 K (J/(g·K)) and depends on the material. Heat capacity is the energy to raise the whole object by 1 K (J/K) and scales with mass: C = m × c.

How many joules is 1 calorie?

The thermochemical calorie is defined as 1 cal = 4.184 J, and the International Table calorie as 4.1868 J. Textbooks often use 4.18 J and introductory courses 4.2 J; use the value stated in the problem.

Should I use K or °C for the temperature change?

A difference is the same in both: from 20 °C to 40 °C is 20 K. Match the unit to the specific heat (J/(g·K) with K, J/(g·°C) with °C). Absolute temperatures require adding 273.15.

Is the mixing temperature just the average?

Only when the masses and specific heats are equal. In general it is a weighted average using the heat capacities m × c, so the larger or higher-capacity object pulls the result towards its temperature.

How do I calculate the heat to melt ice?

While ice at 0 °C becomes water at 0 °C the temperature does not change, so use Q = mL, not Q = mcΔT. Water fusion is about 334 J/g, so 100 g needs about 33,400 J. If the ice is below 0 °C, add the warming step first.

My result does not match the experiment. What should I check?

Check heat losses, the heat capacity of the container, the specific-heat value, how mass and temperature were measured, and the unit conversions. Even with perfect arithmetic the insulated assumption means real heating needs more energy.

How this guide was checked

On 5 October 2026 the definitions and formulas were checked against the sources above, and the heat, specific heat, heat capacity, equilibrium temperature, latent heat and staged calculation examples were recalculated by hand (8,360 J, 8,400 J, 0.45 J/(g·K), 16,800 J, 40.0 °C, 27.78 °C, 33,400 J, 225,600 J and 35,490 J). The datasets are teaching examples, not measured results. The calculator inputs and outputs described here (three modes, presets, unit conversion, working) were also verified.