How to Calculate Heat Transfer

The one equation that runs calorimetry

Coffee-cup calorimetry, bomb calorimetry, food-energy calculations, the warmth of your hand on a metal railing — all of it runs on:

q = m · c · ΔT

where q is heat (J or kJ), m is mass (g), c is specific heat capacity (J/g·°C), and ΔT = T_final − T_initial. Positive q means the substance absorbed heat. Negative q means it released heat. That sign convention is the bookkeeping that lets you balance heat lost by one body against heat gained by another.

The reason a single equation does so much work: as long as you’re not crossing a phase boundary, the heat needed to change a substance’s temperature is just mass times heat capacity times the temperature change. Phase changes (melting, boiling) require their own equations because temperature stays constant while energy goes into rearranging molecules instead of speeding them up.

Specific heats worth memorizing

A handful show up over and over:

Substance	c (J/g·°C)
Water (liquid)	4.184
Ice	2.09
Steam	2.01
Aluminum	0.897
Iron	0.449
Copper	0.385
Silver	0.235
Gold	0.129
Ethanol	2.44

Water’s value (4.184) is anomalously high — that’s the whole reason oceans moderate climate, the body uses sweat to cool, and water makes a great calorimeter fluid. Most metals sit below 1 J/g·°C, which is why a small piece of hot iron cools quickly when dropped into water.

Worked example 1 — heating water

How much heat raises 250.0 g of water from 20.0 °C to 80.0 °C?

ΔT = 80.0 − 20.0 = 60.0 °C q = (250.0)(4.184)(60.0) = 62,760 J = 62.8 kJ

Roughly the energy in a small cookie, all going into 250 mL of water.

Worked example 2 — cooling iron

A 50.0 g piece of iron at 200.0 °C cools to 25.0 °C in water. How much heat did the iron release?

ΔT = 25.0 − 200.0 = −175.0 °C (negative because it cooled) q = (50.0)(0.449)(−175.0) = −3,929 J = −3.93 kJ

Negative q signals heat leaving the iron. That same 3.93 kJ was absorbed by the water around it, which is the basis for the next problem.

Worked example 3 — mixing problem (find T_final)

100.0 g copper at 250.0 °C dropped into 200.0 g water at 22.0 °C. No heat lost to the cup or air. What’s the final temperature?

The setup is conservation of heat: q_copper + q_water = 0.

m_Cu · c_Cu · (T_f − 250.0) + m_water · c_water · (T_f − 22.0) = 0 (100.0)(0.385)(T_f − 250.0) + (200.0)(4.184)(T_f − 22.0) = 0 38.5(T_f − 250.0) + 836.8(T_f − 22.0) = 0 38.5·T_f − 9625 + 836.8·T_f − 18,410 = 0 875.3·T_f = 28,035 T_f = 32.0 °C

Notice the final temperature lands much closer to the water’s starting value than the copper’s. That asymmetry comes from water’s enormous heat capacity combined with twice the mass — the water has roughly 22× the thermal inertia of the copper, so it barely budges.

Worked example 4 — finding specific heat of an unknown metal

A 75.0 g unknown metal at 100.0 °C goes into 150.0 g water at 24.0 °C. Final temperature: 28.8 °C. Specific heat?

Heat gained by water: q_water = (150.0)(4.184)(28.8 − 24.0) = (150.0)(4.184)(4.8) = 3,012 J

Heat lost by metal: q_metal = −3,012 J ΔT_metal = 28.8 − 100.0 = −71.2 °C c_metal = q / (m · ΔT) = −3,012 / [(75.0)(−71.2)] = 0.564 J/g·°C

Close to stainless steel. This is the standard “identify the unknown metal” experiment in introductory labs.

Worked example 5 — food calorimetry

A 2.50 g snack burned in a calorimeter heats 1000 g of water from 21.0 °C to 29.5 °C. Energy per gram?

q = (1000)(4.184)(8.5) = 35,564 J Per gram: 35,564 / 2.50 = 14,226 J/g = 14.2 kJ/g

In food Calories (kcal): 14,226 / 4184 = 3.40 Cal/g. That’s roughly the energy density of carbohydrate, which lines up with most snack foods.

Traps that catch students

ΔT direction. Always T_final − T_initial. Reverse the subtraction and the sign of q flips, which inverts whether the process looks endothermic or exothermic.

Joules vs. kilojoules. Specific heats are in J/g·°C, so q comes out in J. Divide by 1000 to report kJ. Many textbook answer keys use kJ — get the magnitude right by checking which unit the answer expects.

Calorimeter heat capacity. Coffee-cup problems treat the cup as having negligible heat capacity. Bomb calorimetry doesn’t — the calorimeter itself absorbs heat (q = C_cal · ΔT, where C_cal is calibrated for the whole apparatus). Read the problem to know which regime you’re in.

ΔT in °C vs. K. A 10 °C change equals a 10 K change. The two scales have the same size degrees, so ΔT is identical in either unit. You only need 273.15 for absolute temperatures, never for differences.

Confusing solution mass with solvent mass in dilute solutions. Most coffee-cup problems use the water mass (or total dilute solution mass) and the specific heat of water. Concentrated solutions need their own measured c.

Run any of these through the Thermochemistry Calculator to see the step-by-step.