Thermal Equilibrium Simulator

Physics Thermodynamics • Temperature and Zeroth Law

Written by STEM Calculators Team Published January 2, 2026 Updated February 24, 2026

Thermal Equilibrium Simulator

Insulated thermal equilibrium with $Q=mc\Delta T$ and $$Q=mL$$ . The solver automatically applies the correct stage method (warm/cool to phase point, melt/freeze/boil/condense possibly partially, then continue).

Inputs support: pi, e, sqrt( ), sin, cos, exp, log. Use *. Temperatures are in °C. (Constants assume ~1 atm; real values vary with pressure/purity.)

Solve mode

Mode: Unknown: Target $T_f$ (°C):

In mass-solve mode the calculator uses the same stage method to compute per-kg heat for each body up to the given $T_f$, then solves $$m_1 q_1(T_f) + m_2 q_2(T_f)=0$$ .

Body 1

T: — • Q: —

Mass $m_1$ (kg): Initial $T_1$ (°C): Material preset: Thermal model: Specific heat $c_1$ (J/kg·K): Enable melt/freeze: Melting/freezing $T_f$ (°C): Latent heat $L_f$ (J/kg): $c_s$ solid (J/kg·K): $c_\ell$ liquid (J/kg·K): Enable boil/condense: Boiling $T_b$ (°C): Latent heat $L_v$ (J/kg): $c_g$ gas (J/kg·K):

Body 2

T: — • Q: —

Mass $m_2$ (kg): Initial $T_2$ (°C): Material preset: Thermal model: Specific heat $c_2$ (J/kg·K): Enable melt/freeze: Melting/freezing $T_f$ (°C): Latent heat $L_f$ (J/kg): $c_s$ solid (J/kg·K): $c_\ell$ liquid (J/kg·K): Enable boil/condense: Boiling $T_b$ (°C): Latent heat $L_v$ (J/kg): $c_g$ gas (J/kg·K):

Animation + plot settings (optional)

Playback duration (seconds): Curve samples:

The x-axis uses a normalized progress parameter $\tau\in[0,1]$ (not physical time). Drag to pan, mouse wheel to zoom, double-click to reset.

Ready

Steps

Enter values and click Solve.

Animated temperature plot

$T_1(\tau)$, $T_2(\tau)$ → $T_f$ (flat segments appear during phase change)

Scrub $\tau$: τ = 0

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Thermal Equilibrium Simulator — Theory

This tool models insulated heat exchange using only the classic heat-transfer formulas $Q=mc\Delta T$ and $$Q=mL$$ , including automatic detection of phase changes (possibly partial) until thermal equilibrium is reached.

1. Zeroth law and equilibrium

In an insulated system with two bodies exchanging heat, thermal equilibrium occurs when both reach the same final temperature $T_f$. If there is no heat loss to the environment, conservation of energy gives:

\[ Q_1(T_f)+Q_2(T_f)=0 \]

Convention: $Q>0$ means the body absorbs heat; $Q<0$ means it releases heat.

2. Sensible heat

When the material stays in the same phase, heating/cooling is computed by:

\[ Q = mc\Delta T = mc(T_f - T_i) \]

The tool uses the appropriate $c$ for the current phase (solid/liquid/gas) when phase data is enabled.

3. Phase change (latent heat)

During melting/freezing or boiling/condensation, temperature stays constant at the phase point. The heat exchanged is:

\[ Q = mL \]

If the phase change is only partial, the tool uses a fraction $0\le \alpha \le 1$:

\[ Q = mL\,\alpha \]

This is exactly what happens when equilibrium is reached at the melting/freezing temperature (e.g., $0^\circ$C for water): extra heat goes into melting (or comes from freezing) without changing temperature.

4. The stage method (what the calculator does)

To compute $Q_i(T)$ for a body going from $T_i$ to some candidate $T$, the tool applies the same logic you described:

If the body must move toward a phase point (e.g., ice at $-20^\circ$C warming to $0^\circ$C), it uses $Q=mc\Delta T$.
If it crosses a phase point, it applies $Q=mL$ at constant temperature (possibly partial).
Then it continues in the new phase using $Q=mc\Delta T$ until reaching $T$.

The equilibrium temperature $T_f$ is found by enforcing $Q_1(T_f)+Q_2(T_f)=0$, including the possibility that equilibrium occurs exactly at a phase point with a partial phase change.

5. The two classic outcomes for hot water + ice

Case A — hot water dominates (ice melts, final $T_f>0$)

Ice at $T_i<0$ can undergo three stages:

\[ Q_{\text{ice}} = m c_s(0-T_i) + mL_f + m c_\ell(T_f-0) \]

Hot water only cools down:

\[ Q_{\text{water}} = m c_\ell(T_f - T_{i,\text{hot}}) \]

Case B — ice dominates (water may freeze, final $T_f\le 0$)

Hot water cools to $0^\circ$C, then can freeze partially/completely at $0^\circ$C, then the formed ice can cool below $0^\circ$C:

\[ Q_{\text{water}} = m c_\ell(0-T_{i,\text{hot}}) - mL_f\,\alpha + m c_s(T_f-0)\quad (T_f<0) \]

The calculator decides automatically whether $\alpha$ is $0$, partial $0<\alpha<1$, or $1$, based on the energy balance.

6. Unknown mass mode

If one mass is unknown but a target equilibrium temperature $T_f$ is given, the tool computes the heat per kg required for each body (including the same stage method and phase changes), then solves:

\[ m_1 q_1(T_f) + m_2 q_2(T_f) = 0 \]

For example, if $m_1$ is unknown:

\[ m_1 = -\,m_2\,\frac{q_2(T_f)}{q_1(T_f)} \]

The solution is physically valid only if the resulting mass is positive.

7. Notes and limitations

Constants are typical values and depend on pressure/purity; boiling behavior especially can differ.
The model assumes no heat loss to the environment and no mechanical work.
Temperatures below absolute zero ($-273.15^\circ$C) are rejected.
“Olive oil” phase data is approximate and highly composition-dependent (used for qualitative exploration).

Steps

Animated temperature plot

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