Heat Transfer Calculator (Q = mcΔT)

Use Q = mcΔT, where m is mass in kilograms, c is specific heat capacity in J/(kg·K), and ΔT is the temperature change in kelvin or Celsius (the size of one degree is the same in both). Q comes out in joules. Heating 2 kg of water (c = 4186 J/kg·K) by 30 °C needs Q = 2 × 4186 × 30 ≈ 251,000 J. That's a lot, which is why water is such a good heat sink. The formula is your starting point for almost every calorimetry problem and applies wherever there's no phase change.

Rearrange Q = mcΔT to find the specific heat capacity: c = Q/(mΔT). Provide the heat absorbed, the mass, and the temperature change. So if 2000 J raises 0.5 kg of an unknown metal by 20 °C, then c = 2000/(0.5 × 20) = 200 J/(kg·K). That value is consistent with lead. This kind of reverse calculation is how labs identify materials from calorimetry experiments, or how engineers spec coolants. Make sure mass and temperature change use consistent units; mixing kelvin and Celsius for ΔT is fine, but never mix grams and kilograms without converting.

Solve Q = mcΔT for ΔT, then add to the starting temperature: T_f = T_i + Q/(mc). Pour 50000 J into 1 kg of water already at 20 °C, and T_f = 20 + 50000/(1 × 4186) ≈ 31.9 °C. This works as long as no phase change happens — the moment water hits 100 °C and starts boiling, you need Q = mL for the latent heat instead. So always sanity-check that your calculated final temperature stays within a single phase. Otherwise, split the problem into stages.

Rearrange to m = Q/(cΔT) when you know how much heat was supplied, the substance's specific heat, and the temperature change. If 84,000 J heated some water by 10 °C, then m = 84000/(4186 × 10) ≈ 2.0 kg. This is the kind of question that comes up in industrial heating calculations or when you're trying to figure out how much coolant a system needs. The same algebra works for cooling, just with the heat removed and ΔT being negative. Keep units consistent and the formula doesn't lie.

The SI unit is J/(kg·K), since heat is in joules, mass in kilograms, and temperature change in kelvin. Some textbooks use J/(g·°C), which is just 1000 times smaller for the same number — water is 4.186 J/(g·°C) but 4186 J/(kg·K). Older engineering texts may show calories per gram per degree, where 1 cal = 4.184 J. Mixing these up is the single most common error in calorimetry. Pick a unit system and stick with it. The size of a degree change is identical in kelvin and Celsius, so ΔT doesn't care.

No, Q = mcΔT only handles temperature change within a single phase — solid, liquid, or gas. During a phase change, temperature stays fixed while heat goes into rearranging molecular bonds, so you use Q = mL instead, where L is the latent heat (of fusion or vaporisation). Heating ice from −10 °C to steam at 110 °C needs four stages: warming the ice, melting it, warming the water, boiling it, and warming the steam. Three of those use Q = mcΔT, two use Q = mL. Always sketch the heating curve before plugging numbers.

1 calorie equals 4.184 joules, and 1 kilocalorie (the 'food calorie') equals 4184 joules. So a 200 kcal snack delivers about 836,800 J of energy if fully metabolised. Many older textbooks or nutrition contexts give heat in calories, but SI calculations work in joules, so convert before plugging into Q = mcΔT. Calculators often have a unit picker, but doing the conversion once by hand cements the relationship. Just remember that food labels mean kilocalories even when they say 'calories' — that thousandfold difference matters when you're checking dimensions.