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Entropy Change Calculator

Find the entropy change of a process.

How it is calculated

For heat transferred reversibly at a constant temperature, the entropy change is ΔS = Q / T. Enter the heat Q in joules and the absolute temperature T in kelvin; the answer is in J/K with four decimals. Heat absorbed by the system is positive and gives a positive ΔS; heat released is negative. Temperature must be above zero, because it is on the kelvin scale. The formula only holds when T stays the same during the transfer, as in melting, boiling or a large heat reservoir. For a changing temperature you need to integrate dQ / T instead.

Worked example

A system absorbs Q = 5,000 J of heat at a constant T = 300 K. ΔS = 5,000 / 300 = 16.6667 J/K. The entropy change is 16.6667 J/K. The value is positive because heat flows in. If the same 5,000 J were released at 300 K, the result would be −16.6667 J/K.

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Common mistakes

  • Entering temperature in °C. The formula needs kelvin; add 273.15 to a Celsius value.
  • Entering heat in kilojoules or calories while reading the answer as J/K.
  • Using the formula when temperature changes during heating, where Q / T with a single T is wrong.
  • Dropping the sign. Heat leaving the system must be entered as negative.
  • Applying it to the whole universe without thinking: the entropy change of a system alone can be negative.

About this calculator

This calculator finds the entropy change of an isothermal, reversible process using ΔS = Q/T, where Q is the heat transferred and T is the absolute temperature (in kelvin) at which the transfer happens. Entropy is the thermodynamic measure of disorder or the number of microscopic ways energy can be arranged, and the second law of thermodynamics states that the total entropy of an isolated system never decreases.

Chemists and physicists use this formula to analyze phase changes like melting and boiling (where heat is absorbed or released at a constant temperature), to evaluate the efficiency limits of heat engines and refrigeration cycles, and to determine whether a process is thermodynamically favorable. Chemistry and physics students, thermodynamics researchers, and mechanical and chemical engineers designing engines or industrial processes use it to quantify how much a system's disorder increases or decreases.

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Comments (1)

  • Editorial team

    Melting and boiling are the cases where this formula fits best, because the temperature really does stay fixed. Melting 1 kg of ice takes roughly 334,000 J at 273.15 K, which gives about 1,222.77 J/K for the water. Freezing the same mass is the same figure with a negative sign. Which phase change or heat-reservoir problem are you working on? Share the values you entered and what you expected to see.

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