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❄️ Sensible Heat Calculator

Calculate the sensible heat load required to change the temperature of a substance without changing its phase.

❄️ Sensible Heat Calculator

Calculate the sensible heat load required to change the temperature of a substance without changing its phase.

✅ Calculation Result

Sensible Heat Load (Q)-BTU/s

Heating/Cooling Process

20°C
T₁
❓
35°C
T₂
FLOW ➔

Calculator Description

Sensible heat is the heat exchanged when a substance changes temperature without changing phase. Unlike latent heat (melting, boiling), it is the heat that makes a thermometer reading rise or fall during heating or cooling.

What this calculator finds

Given a mass flow rate and specific heat together with the initial and final temperatures, it computes the required sensible-heat load Q. It is used for HVAC heating/cooling loads and process heating/cooling energy.

Why it matters

  • Sizing HVAC heating/cooling loads and coil capacity
  • Estimating heating/cooling energy for heat exchangers, boilers and chillers
  • Evaluating heat duty for temperature control of process streams

Formula

Sensible Heat Formula

It is the mass flow rate m times the specific heat C_p times the temperature change ΔT (= T₂ − T₁). In metric, kg/s × kJ/(kg·K) × K = kJ/s = kW; in imperial, lb/s × BTU/(lb·°F) × °F = BTU/s.

Q=mCp(T2−T1)Q = m C_p\left(T_2 - T_1\right)
  • Q — Sensible heat load (heat rate) [kW or BTU/s]
  • m — Mass flow rate [kg/s or lb/s]
  • C_p — Specific heat [kJ/(kg·K) or BTU/(lb·°F)]
  • T₁ — Initial temperature [°C or °F]
  • T₂ — Final temperature [°C or °F]
  • ΔT — Temperature change (T₂ − T₁) [K or °F]

How the formula works

  • Larger mass flow or higher specific heat needs more heat for the same temperature change.
  • Load grows linearly with ΔT. If T₂ < T₁, Q is negative, meaning cooling (heat removal).
  • Specific heat varies with temperature and substance; use the operating-temperature value for precision (air ≈ 1.006 kJ/(kg·K)).

Volumetric-flow form

If density ρ is known, substitute m = ρ·V (volumetric flow) to get Q = ρ·V·C_p·ΔT, a form common in HVAC load calculations.

Worked example

Heating air at m = 1.5 kg/s from 20 °C to 35 °C with C_p = 1.006 kJ/(kg·K) and ΔT = 15 K needs Q = 1.5 × 1.006 × 15 ≈ 22.6 kW. Cooling (35→20 °C) gives Q ≈ −22.6 kW (heat removed).

Useful Tips

Practical tips

  • In HVAC loads, separate and sum sensible and latent (moist-air) heat.
  • Specific heat varies widely by substance (water ≈ 4.18, air ≈ 1.0); wrong C_p throws the load off by factors.
  • For batch (total mass) heating rather than continuous flow, divide by time to get the load rate.

Limitations & cautions

  • If phase change (evaporation/condensation) occurs, add latent heat — this formula alone is insufficient.
  • Treating C_p as constant ignores its temperature dependence, causing error at high temperature ranges.
  • Heat losses (convection, radiation) must be added separately; this is an ideal adiabatic heating/cooling relation.