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Generated on: July 20, 2026 at 09:33 PM

Specific Heat Calculator

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Mass1 kg
Specific Heat Capacity (J/kg·K)4,186
Initial Temperature20 °C
Final Temperature100 °C

Generated with OneCalc · https://onecalc.app/en/physics/specific-heat

Physics

Specific Heat Calculator

Calculate the heat energy Q = mcΔT needed to change a substance’s temperature, from its mass, specific heat capacity, and temperature change. Results in J, kJ, cal, or kcal.

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Specific heat capacity explained

Specific heat capacity is the amount of energy required to raise the temperature of one kilogram of a substance by one kelvin. The heat QQ exchanged when a body of mass mm and specific heat capacity cc undergoes a temperature change ΔT\Delta T is given by Q=mcΔTQ = mc\Delta T. A positive QQ indicates heat absorbed; a negative QQ indicates heat released.

The specific heat formula

Q=m×c×ΔTQ = m \times c \times \Delta T
SymbolQuantitySI unit
QHeat energyJ (joules)
mMasskg
cSpecific heat capacityJ/(kg·K)
ΔTTemperature change (T_final − T_initial)K (= °C numerically)

A positive Q means the substance absorbed heat (temperature rose). A negative Q means it released heat (temperature fell). A temperature difference in Kelvin and Celsius is numerically identical, so the formula works with either scale for ΔT.

Specific heat capacity values

Materialc (J/kg·K)Notes
Water (liquid)4,186Highest of common substances
Ice2,090
Steam2,010
Ethanol2,440
Aluminum900Common in cookware
Air1,005At constant pressure
Iron449
Copper385
Lead128Lowest common metal
Glass~840Varies by composition

Water's exceptionally high specific heat (4,186 J/kg·K) is why it takes so long to heat a pot of water — and why the ocean moderates coastal climates.

Worked example

Heating 2 kg of water from 15 °C (288.15 K) to 95 °C (368.15 K):

ΔT=95−15=80 °C=80 K\Delta T = 95 - 15 = 80\text{ °C} = 80\text{ K} Q=2 kg×4,186 J/(kg\cdotpK)×80 K=669,760 J≈670 kJQ = 2\text{ kg} \times 4{,}186\text{ J/(kg·K)} \times 80\text{ K} = 669{,}760\text{ J} \approx 670\text{ kJ}

One dietary kilocalorie (food calorie, kcal) = 4,184 J, so this equals about 160 kcal — roughly the food energy in a banana.

Heating 5 lb of water from 60 °F to 200 °F, using water's specific heat of 1 BTU/(lb·°F):

ΔT=200−60=140 °F\Delta T = 200 - 60 = 140\text{ °F} Q=5 lb×1 BTU/(lb\cdotp°F)×140 °F=700 BTUQ = 5\text{ lb} \times 1\text{ BTU/(lb·°F)} \times 140\text{ °F} = 700\text{ BTU}

By definition, 1 BTU raises 1 lb of water by 1 °F, so this is the energy a typical kitchen stove burner delivers in roughly 30 seconds.

Why water has such high specific heat

Water molecules form hydrogen bonds that require significant energy to disrupt before temperature increases. This is an emergent property of water's molecular structure that makes it unusual among liquids. Practical consequences:

  • Climate regulation: Coastal cities have smaller temperature swings than inland cities at the same latitude.
  • Cooking: Water absorbs large amounts of heat before boiling, making it ideal for even cooking.
  • Cooling systems: Car radiators, industrial heat exchangers, and CPU coolers all exploit water's heat capacity.

Calorimetry

Calorimetry measures heat exchange between substances. In a coffee-cup calorimeter experiment:

  1. A hot metal sample (mass m₁, specific heat c₁) is dropped into cooler water (mass m₂, c₂ = 4,186 J/kg·K).
  2. The mixture reaches thermal equilibrium at temperature T_eq.
  3. Heat lost by metal = Heat gained by water:
m1c1(T1−Teq)=m2c2(Teq−T2)m_1 c_1 (T_1 - T_{eq}) = m_2 c_2 (T_{eq} - T_2)

Solving for c₁ gives the unknown specific heat. This is the standard measurement technique in high school and university chemistry labs.

Frequently Asked Questions (FAQ)

What is the formula for heat transfer?

The heat transfer formula is Q = m × c × ΔT, where Q is heat energy (joules), m is mass (kg), c is specific heat capacity (J/kg·K), and ΔT is the temperature change (T_final − T_initial, in K or °C — numerically identical for a difference). A positive Q means heat is absorbed (heating); a negative Q means heat is released (cooling).

What is the specific heat capacity of water?

Water has a specific heat capacity of approximately 4,186 J/(kg·K) — one of the highest of any common substance. This means it takes 4,186 joules to raise 1 kg of water by 1 °C. By comparison: aluminum is ~900 J/(kg·K), copper ~385 J/(kg·K), iron ~449 J/(kg·K), and air ~1,005 J/(kg·K). Water's high specific heat is why it is used as a coolant and why coastal climates are milder than inland ones.

What does a positive or negative Q value mean?

Q > 0 means the object absorbs heat (its temperature rises). Q < 0 means the object releases heat (its temperature falls). For example, heating 1 kg of water from 20 °C to 100 °C gives Q = +334,880 J (heat absorbed). Cooling the same water back to 20 °C gives Q = −334,880 J (heat released to the surroundings).

How is specific heat used in calorimetry experiments?

In a calorimetry experiment, a hot object is placed in a cooler liquid inside an insulated cup. Heat lost by the hot object (Q = m₁c₁ΔT₁) equals heat gained by the liquid (Q = m₂c₂ΔT₂) when the system reaches equilibrium. By measuring masses and temperature changes, you can calculate an unknown specific heat. This method is standard in high school and university chemistry labs.

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