Entropy change calculator

What entropy change tells you

Entropy is a property of the state of a system, like pressure or internal energy, so ΔS depends only on the starting and ending states and not on the path between them. For heat added reversibly at one temperature, OpenStax University Physics (Volume 2, section 4.6) defines ΔS = Q ÷ T. The second law then says the total entropy of a system and everything it exchanges heat with never decreases, ΔS ≥ 0, with equality only for a reversible process. A positive total means the process can happen by itself and cannot be undone for free; a negative total means it cannot happen on its own. This calculator gives the entropy change of the thing you are heating, melting or mixing and, if you enter a surroundings temperature, the total and the verdict.

The formulas

Heat Q at a constant temperature T (kelvins): ΔS = Q ÷ T. Melting or boiling is such a case, with Q = mL and T the melting or boiling point (Example 4.4). Warming or cooling a mass m with specific heat c: ΔS = m·c·ln(T2 ÷ T1), because dQ = m·c·dT and the integral of dT ÷ T is a logarithm (Example 4.5). An ideal gas between two states: ΔS = nR ln(V2 ÷ V1) + n·Cv·ln(T2 ÷ T1), with R = 8.314 J/(mol·K). Surroundings that behave as a large reservoir at T_R and give the system the heat Q change by −Q ÷ T_R. Two samples mixing without losses end at Tf = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂), and each changes by m·c·ln(Tf ÷ T).

Worked example: 50 g of ice melting in a 20°C room

The total is positive, so the second law allows it and the process is irreversible. OpenStax Example 4.7 gets 76.3, −71.7 and 4.6 J/K because it uses L_f = 335 J/g and c = 4.19 J/(g·K) and rounds the heat to 2.10 × 10⁴ J; the difference is about half a percent and the conclusion is the same.

Common mistakes

Using degrees Celsius or Fahrenheit in Q ÷ T or in the logarithm gives nonsense; the unit menu converts for you, but a value typed in the wrong scale is still wrong. Forgetting the surroundings is the second most common error: a cooling object has a negative ΔS and that is fine as long as the surroundings gain more. Students also think an irreversible process has no entropy change; ΔS of the system is the same for any path between the same two states, so you compute it along a reversible path (OpenStax Example 4.5 does this for a rapidly cooled object). Finally, the sign of Q: it is positive when heat enters the system, so the reservoir always gets the opposite sign.

Limits of this method

Specific heat is treated as constant. The OpenStax table gives one value per material (ice is averaged from −50°C to 0°C, water is quoted at 15°C, and steam at constant pressure), and real values drift by a few percent over a wide temperature range; the specific heat calculator on this site uses NIST values for 25°C and will differ a little. The gas formula assumes an ideal gas with constant Cv, and OpenStax notes that real gases have somewhat higher heat capacities than the table values, especially when molecules can vibrate. The surroundings are assumed to stay at one temperature, and entropy is defined here only between equilibrium states.

How do I find the entropy change of melting ice?

Melting happens at one fixed temperature, so ΔS = Q ÷ T with Q = mL_f (OpenStax University Physics, Volume 2, Example 4.4). For 50 g of ice, Table 1.4 gives L_f = 334 kJ/kg, so Q = 16.7 kJ and ΔS = 16,700 J ÷ 273.15 K = 61.1 J/K. The book prints 61.5 J/K because it uses 335 J/g and rounds Q to 16.8 kJ first; the answers agree to under 1%. Type your own latent heat in the "Melting, freezing, boiling" tab if your course uses a slightly different value.

Why can the entropy of an object go down without breaking the second law?

The second law applies to the total. A cooling object loses entropy, but the surroundings that take its heat gain more. OpenStax asks about a 5.0-kg block of lead at 600°C dropped into a lake at 15°C. With c = 128 J/(kg·K) the lead loses 710 J/K, the lake gains 1,300 J/K, and the universe gains about 590 J/K. The answer key shows 591 J/K because it subtracts the rounded parts. Fill in the surroundings temperature on the first tab and the calculator shows each part and the total.

How do I calculate the entropy change when two objects reach equilibrium?

Find the common final temperature, then add m·c·ln(Tf ÷ T) for each object. OpenStax Example 4.9 takes two 4.00-kg steel objects, c = 450 J/(kg·K), at 400 K and 300 K: the final temperature is 350 K, the hot one changes by −240 J/K, the cold one by +277 J/K, and the total is +37.1 J/K, which the book rounds to 37 J/K. The total is positive for any two different starting temperatures; the "Mixing two samples" tab does this for any materials, masses and temperatures.

How is the entropy change of an ideal gas calculated?

ΔS = nR ln(V2 ÷ V1) + n·Cv·ln(T2 ÷ T1). The first term is the change from the larger volume, the second from the temperature change. OpenStax Table 3.3 gives Cv = 3/2 R = 12.5 J/(mol·K) for a monatomic gas, 5/2 R = 20.8 for a diatomic gas and 3R = 24.9 for a polyatomic gas. In the Stirling engine of Example 4.6 (0.0010 mol of a monatomic gas, 133°C and 33°C), the isothermal expansion from 0.10 to 0.20 m³ gives +0.00576 J/K and the cooling at constant volume gives −0.00353 J/K. A free expansion into a vacuum has no heat flow, yet ΔS = nR ln(V2 ÷ V1) > 0 for the universe (Example 4.8).

Why must the temperature be in kelvins?

Entropy is defined through dQ ÷ T, and T has to be the absolute temperature: the formula needs the ratio of two temperatures, and a ratio of Celsius or Fahrenheit readings has no physical meaning (20°C to 60°C is a ratio of 3 in Celsius but only 1.14 in kelvins). The calculator accepts °F, °C or K and converts to kelvins before it takes the logarithm or divides.

What are the units of entropy, and what is BTU/°R?

Entropy is energy per temperature: joules per kelvin (J/K) in SI. In US customary units it is BTU per degree Rankine: 1 BTU/°R = 1,900 J/K, since a Rankine degree is 5/9 kelvin and 1 BTU (IT) is 1055.06 J. The calculator shows both, and calories per kelvin (1 cal = 4.184 J).

Sources: OpenStax University Physics, Volume 2, sections 4.6 “Entropy” (Examples 4.4 to 4.6) and 4.7 “Entropy on a Microscopic Scale” (Examples 4.7 to 4.9), the chapter 4 problems and answer key, section 3.5 (Table 3.3, molar heat capacities) and sections 1.4 and 1.5 (Tables 1.3 and 1.4, specific and latent heats). Related: Carnot efficiency calculator, latent heat calculator, specific heat calculator and adiabatic process calculator.