Sensible Heat Load
Heat needed to change air or water temperature without a phase change.
Formula
Variables
Example
2 kg water, c=4186, ΔT=10: Q = 83,720 J
How it works
This formula calculates the sensible heat needed to raise or lower the temperature of a substance without changing its phase. It works because the energy required depends on how much material there is, how hard that material is to warm up (its specific heat), and how many degrees its temperature changes. In HVAC and refrigeration, it is used for air, water, and other fluids when only temperature changes matter, not boiling or condensing.
When to use it
Use this when a liquid or gas is being heated or cooled and it stays in the same state, such as warming supply air, cooling chilled water, or estimating room cooling loads. It is not the right formula for melting, freezing, evaporation, or condensation.
Worked example, step by step
A 3 kg metal part is cooled from 80°C to 25°C. Its specific heat is 900 J/kg·°C. Step 1: Find the temperature change. ΔT = 25 - 80 = -55°C Step 2: Use Q = mcΔT. Q = 3 × 900 × (-55) Step 3: Calculate. Q = -148,500 J Final answer: Q = -148,500 J, meaning 148,500 J of heat must be removed from the part.
Common mistakes to avoid
- Using the formula when the substance changes phase, such as water boiling into steam
- Forgetting that ΔT can be negative when the object is cooling
- Mixing up units, especially using grams instead of kilograms or °F instead of °C
Did You Know?
About 60% of a home's energy bill in cold climates goes to sensible heating of indoor air.
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