Specific heat capacity tells you how much energy it takes to raise the temperature of a material by one degree. Water's specific heat is 4.186 J/(g·°C) (or 1.0 BTU/(lb·°F)), which is why it takes so much energy to heat your home's hot water and why water-based HVAC systems are so effective at storing and transferring thermal energy.
The core formula is Q = mcΔT — heat energy equals mass times specific heat times temperature change. Use the calculator below to solve for any unknown variable, or read on for detailed explanations and real-world HVAC applications.
Specific Heat Capacity Calculator
How to Use This Calculator
Select what you want to solve for:
- Heat Energy (Q) — How much energy is needed? Enter mass, specific heat, and temperature change.
- Mass (m) — How much material is involved? Enter heat energy, specific heat, and temperature change.
- Specific Heat (c) — What's the material's heat capacity? Enter heat energy, mass, and temperature change.
- Temperature Change (ΔT) — How much will the temperature change? Enter heat energy, mass, and specific heat.
The Specific Heat Formula Explained
Q = mcΔT
| Variable | Symbol | Units (SI) | Units (Imperial) | What It Means |
|---|---|---|---|---|
| Heat Energy | Q | Joules (J) | BTU | Total energy absorbed or released |
| Mass | m | grams (g) or kg | pounds (lb) | Amount of material |
| Specific Heat | c | J/(g·°C) | BTU/(lb·°F) | Energy needed per unit mass per degree |
| Temperature Change | ΔT | °C | °F | Final temp minus initial temp |
Rearranged forms:
- To find energy: Q = mcΔT
- To find mass: m = Q / (cΔT)
- To find specific heat: c = Q / (mΔT)
- To find temperature change: ΔT = Q / (mc)
Unit Conversions
| Conversion | Factor |
|---|---|
| 1 BTU | 1,055.06 Joules |
| 1 kWh | 3,412 BTU = 3,600,000 Joules |
| 1 calorie | 4.186 Joules |
| 1 therm | 100,000 BTU = 105,506,000 Joules |
| °C to °F temperature change | Multiply ΔT(°C) by 1.8 to get ΔT(°F) |
Specific Heat Values for Common Materials
Fluids Used in HVAC Systems
| Material | Specific Heat (J/(g·°C)) | Specific Heat (BTU/(lb·°F)) | HVAC Application |
|---|---|---|---|
| Water | 4.186 | 1.000 | Hydronic heating, water heaters |
| Propylene Glycol (50%) | 3.400 | 0.812 | Antifreeze loops, solar thermal |
| Ethylene Glycol (50%) | 3.300 | 0.788 | Chiller systems |
| Refrigerant R-410A (liquid) | 1.720 | 0.411 | Air conditioners, heat pumps |
| Refrigerant R-32 (liquid) | 1.940 | 0.463 | Modern heat pumps |
| Air (dry, sea level) | 1.006 | 0.240 | Forced-air HVAC |
| Steam (100°C) | 2.010 | 0.480 | Steam heating systems |
Building Materials
| Material | Specific Heat (J/(g·°C)) | Density (kg/m³) | Thermal Mass Rating |
|---|---|---|---|
| Concrete | 0.880 | 2,400 | Very High |
| Brick | 0.840 | 1,920 | High |
| Granite/Stone | 0.790 | 2,650 | Very High |
| Drywall (gypsum) | 1.090 | 800 | Low |
| Wood (oak) | 2.380 | 700 | Moderate |
| Wood (pine) | 2.300 | 500 | Low-Moderate |
| Steel | 0.490 | 7,850 | Moderate (dense) |
| Aluminum | 0.897 | 2,700 | Moderate |
| Glass | 0.840 | 2,500 | Moderate |
| Fiberglass insulation | 0.700 | 12-50 | Very Low |
| Soil (dry) | 0.800 | 1,500 | Moderate |
| Soil (wet) | 1.480 | 1,800 | High |
Why water dominates HVAC. Water has one of the highest specific heat capacities of any common substance — 4.186 J/(g·°C). This means water can absorb and transport large amounts of heat energy per unit of mass. That's why hydronic (water-based) heating and cooling systems are so efficient: water carries 3,500 times more heat per unit volume than air.
HVAC Applications of Specific Heat
Application 1: Calculating Water Heater Energy Use
Problem: How much energy does it take to heat 50 gallons of water from 55°F (incoming cold water) to 120°F (tank setting)?
Setup:
- Mass: 50 gallons = 417 lbs (8.34 lb/gal)
- Specific heat of water: 1.0 BTU/(lb·°F)
- Temperature change: 120°F - 55°F = 65°F
Calculation: Q = mcΔT = 417 × 1.0 × 65 = 27,105 BTU
Converting to kWh: 27,105 BTU ÷ 3,412 BTU/kWh = 7.94 kWh
Cost at average rate: 7.94 kWh × $0.168/kWh = $1.33 per full tank heat-up
If your family drains and reheats the equivalent of one full tank per day, that's approximately $40/month for water heating — consistent with national averages.
Application 2: Sizing a Hydronic Heating System
Problem: A hydronic baseboard system needs to deliver 40,000 BTU/hr to heat a home. The water enters the baseboard loop at 180°F and returns at 160°F. What flow rate is needed?
Setup:
- Q = 40,000 BTU/hr
- c = 1.0 BTU/(lb·°F)
- ΔT = 180 - 160 = 20°F
Calculation: m = Q / (cΔT) = 40,000 / (1.0 × 20) = 2,000 lb/hr of water
Converting: 2,000 lb/hr ÷ 8.34 lb/gal = 240 gallons per hour = 4.0 GPM
So the circulation pump needs to deliver at least 4 gallons per minute.
Application 3: How Long Does It Take for a House to Cool Down?
Problem: A 2,000 sq ft home with moderate thermal mass. The AC shuts off during a power outage when the interior is 72°F. Outside temperature is 95°F. How quickly does the interior temperature rise?
This is a simplified analysis. The "thermal mass" of the home includes air, furniture, drywall, and structural elements.
Approximate thermal mass for a typical 2,000 sq ft home:
- Air: ~5,000 cubic feet = 380 lbs at standard density. c = 0.24 BTU/(lb·°F). Thermal capacity = 91 BTU/°F
- Drywall: ~5,000 lbs. c = 0.26 BTU/(lb·°F). Thermal capacity = 1,300 BTU/°F
- Furniture/contents: ~3,000 lbs. c ≈ 0.30 BTU/(lb·°F). Thermal capacity = 900 BTU/°F
- Total approximate thermal capacity: ~2,300 BTU/°F
If the home gains about 15,000 BTU/hr through walls, windows, and infiltration with a 23°F temperature differential:
Temperature rise rate: 15,000 BTU/hr ÷ 2,300 BTU/°F = ~6.5°F per hour
So after 1 hour: ~78.5°F. After 2 hours: ~85°F. After 3 hours: ~91°F (approaching outdoor temp).
Homes with higher thermal mass (concrete floors, brick walls, stone countertops) maintain temperature much longer during AC outages or thermostat setbacks. A concrete slab home might gain only 3-4°F per hour compared to 6-7°F for a wood-frame home with the same insulation. This is why pre-cooling strategies work well for TOU rate plans — thermal mass holds the cool.
Application 4: Antifreeze Impact on Solar Thermal Efficiency
Problem: A solar thermal system uses a 50/50 propylene glycol/water mix instead of pure water. How does this affect the system's heat-carrying capacity?
Comparison:
- Pure water: c = 1.0 BTU/(lb·°F), density = 8.34 lb/gal
- 50% propylene glycol: c = 0.812 BTU/(lb·°F), density = 8.80 lb/gal
Heat capacity per gallon:
- Water: 8.34 × 1.0 = 8.34 BTU/(gal·°F)
- Glycol mix: 8.80 × 0.812 = 7.15 BTU/(gal·°F)
The glycol mix carries 14% less heat per gallon than pure water. To compensate, the system needs either a higher flow rate or a larger temperature differential. This is why solar thermal systems with antifreeze are typically designed with slightly larger heat exchangers.
Step-by-Step Examples
Example 1: Energy to Heat a Swimming Pool
Problem: How much energy to heat a 15,000-gallon pool from 65°F to 82°F?
Calculation:
- Mass: 15,000 gal × 8.34 lb/gal = 125,100 lbs
- c = 1.0 BTU/(lb·°F)
- ΔT = 82 - 65 = 17°F
- Q = 125,100 × 1.0 × 17 = 2,126,700 BTU
Converting: 2,126,700 BTU ÷ 3,412 = 623 kWh
At $0.168/kWh: $104.67 to heat the pool (electricity only)
With a heat pump pool heater (COP 5.0): 623 kWh ÷ 5.0 = 124.6 kWh × $0.168 = $20.93 — about 80% cheaper.
Example 2: Energy Cost of a Hot Shower
Problem: A 10-minute shower at 2.0 GPM using 105°F water when inlet temp is 55°F.
Calculation:
- Volume: 10 min × 2.0 GPM = 20 gallons
- Mass: 20 × 8.34 = 166.8 lbs
- ΔT: 105 - 55 = 50°F
- Q = 166.8 × 1.0 × 50 = 8,340 BTU = 2.44 kWh
At $0.168/kWh (electric tank, 92% efficient): 2.44 ÷ 0.92 × $0.168 = $0.45 per shower
For a family of four showering daily: $0.45 × 4 × 365 = $657/year on shower water heating alone.
With a low-flow 1.5 GPM showerhead: 15 gallons, 6,255 BTU, 1.83 kWh, $0.33/shower, $482/year. Savings: $175/year from a $15 showerhead.
Example 3: Thermal Storage in a Concrete Floor
Problem: A 1,500 sq ft concrete slab floor (4 inches thick). How much energy can it store if heated from 68°F to 78°F using radiant floor heating?
Calculation:
- Volume: 1,500 sq ft × (4/12) ft = 500 cubic feet
- Mass: 500 ft³ × 150 lb/ft³ = 75,000 lbs
- c of concrete = 0.21 BTU/(lb·°F)
- ΔT = 78 - 68 = 10°F
- Q = 75,000 × 0.21 × 10 = 157,500 BTU = 46.2 kWh
That's equivalent to about 4-5 hours of heating for a typical 2,000 sq ft home. This stored energy is released gradually as the slab cools, providing comfort even after the heating system cycles off — a natural thermal battery.
Advanced Concepts
Volumetric Heat Capacity
For HVAC applications, volumetric heat capacity (energy per unit volume per degree) is often more useful than mass-based specific heat, because we pump fluids by volume, not by weight.
| Material | Volumetric Heat Capacity (kJ/(L·°C)) | vs. Water |
|---|---|---|
| Water | 4.186 | 1.00x (baseline) |
| 50% Propylene Glycol | 3.42 | 0.82x |
| Concrete | 2.11 | 0.50x |
| Air | 0.0012 | 0.0003x |
Water stores 3,500 times more heat per liter than air. This is why hydronic systems use small pipes (3/4 inch to 1 inch) while forced-air systems need large ducts (6 to 14 inches) to move equivalent amounts of heat.
Phase Change and Latent Heat
The Q = mcΔT formula only applies when a material stays in the same phase (liquid water stays liquid, for example). When materials change phase (ice melts to water, water boils to steam), additional "latent heat" energy is required that doesn't change the temperature.
| Phase Change | Latent Heat (Water) | BTU Equivalent |
|---|---|---|
| Melting (ice → water at 32°F) | 334 J/g | 144 BTU/lb |
| Vaporization (water → steam at 212°F) | 2,260 J/g | 970 BTU/lb |
This is relevant to HVAC because refrigerants work by changing phase. When R-410A evaporates inside your indoor coil, it absorbs enormous amounts of heat from your home's air. When it condenses in the outdoor coil, it releases that heat outdoors. The phase change is what makes heat pumps and air conditioners so effective.
Key Takeaways:
- Q = mcΔT is the fundamental formula for heat transfer calculations
- Water has one of the highest specific heats (4.186 J/(g·°C) or 1.0 BTU/(lb·°F))
- Water carries 3,500x more heat per volume than air — explaining hydronic system efficiency
- Heating 50 gallons of water from 55°F to 120°F requires about 8 kWh ($1.33)
- Higher thermal mass in your home (concrete, brick) maintains temperature longer
- Antifreeze solutions carry 14-19% less heat than pure water per gallon
- Phase changes (evaporation, condensation) are the engine behind heat pumps and AC systems
Frequently Asked Questions
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