Mixing Hot and Cold Water: Find Final Temperature from an Energy Balance

For two water samples with the same assumed constant specific heat, no phase change and negligible heat loss or container heat capacity, final temperature is Tf = (mhotThot + mcoldTcold)/(mhot + mcold).

Write the assumptions beside the equation

The weighted-temperature result is an energy balance, not a rule for every pair of hot and cold materials. It assumes both samples remain liquid, share the same specific heat in the model, and exchange no net heat with a container or surroundings. Other materials or phase changes require a broader balance.

Start with heat lost by the hotter water equal to heat gained by the colder water: mhot c(Thot − Tf) = mcold c(Tf − Tcold). The common c cancels. Expanding and gathering the Tf terms gives the displayed weighted formula.

Worked mixture

Mix 0.20 kg of water at 80°C with 0.30 kg at 20°C in the ideal insulated model. The final temperature is [0.20(80) + 0.30(20)]/(0.20 + 0.30) = (16 + 6)/0.50 = 44°C.

Check both sides of the balance without choosing a numerical c. The hot side gives 0.20c(80 − 44) = 7.2c. The cold side gives 0.30c(44 − 20) = 7.2c. Equal values confirm the result, and 44°C lies between the two initial temperatures.

Why a simple average fails here

The unweighted average of 80 and 20 is 50°C, which would be correct for equal water masses under these assumptions. Here the colder sample has greater mass, so the final value lies closer to 20°C. The mass-weighted calculation represents that greater thermal contribution.

Using Celsius is convenient because the derivation uses temperature differences and a weighted average with weights summing to one. Kelvin gives the same physical final state when all temperatures are converted consistently. Do not mix Celsius and Kelvin inputs in one substitution.

Common mistake

If the calculated final temperature lies outside the initial range in this passive two-water model, recheck the signs, masses and assumptions. A container that absorbs heat, heat exchange with the room or an added phase change can alter the answer. These are hypothetical calculations, not instructions to prepare liquids for drinking, medical care or experiments.

Heat lost aur heat gained barabar likho; mass alag ho to plain average mat lo.

Three practice questions with worked answers

1. Equal masses of water at 60°C and 20°C mix ideally. Find Tf.

   Equal masses give their simple average: (60 + 20)/2 = 40°C.

2. Mix 0.10 kg at 90°C and 0.20 kg at 30°C under the model. Find Tf.

   Tf = [0.10(90) + 0.20(30)]/0.30 = 15/0.30 = 50°C.

3. Mix 0.30 kg at 70°C and 0.10 kg at 10°C ideally. Find Tf.

   Tf = [0.30(70) + 0.10(10)]/0.40 = 22/0.40 = 55°C.

Related study guides

• Heat and Temperature: Specific Heat Numericals with a Simple Method
• Thermodynamics for JEE and NEET: Formulas, Sign Conventions and Solved Problems

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Temperature Change and Heat Capacity. No textbook or exam-board affiliation is claimed.