Calorimetry: Measuring Heat with q = mcΔT

Key Ideas

• Specific heat capacity (c): heat required to change 1 g of a substance by 1 °C (or 1 K). Units: J/(g·°C). Water's is unusually high: 4.184 J/(g·°C).
• Heat capacity (C): heat required to change an entire object (a calorimeter, a block of metal) by 1 °C. Units: J/°C. Related by C = mc.
• ΔT is a state change in temperature. A positive ΔT means the substance gained heat; a negative ΔT means it lost heat.
• The sign of q follows the substance. q > 0 means the substance absorbed heat; q < 0 means it released heat.
• Heat flows from hot to cold. A hot metal dropped in cool water cools down (q_metal < 0) while the water warms up (q_water > 0), with q_metal = −q_water.
• 1 °C = 1 K for temperature changes, so c has the same value in °C or K.

Equations and Variables

• q = mcΔT
• q = CΔT (where C = mc)
• q_system = −q_surroundings

How It Works

A calorimeter is an insulated container that prevents heat from escaping to the room. A known process (a reaction, a hot object, an electrical heater) transfers heat to or from a known mass of water (or the calorimeter itself), and the temperature change is measured. The heat is then computed with q = mcΔT.

There are two standard devices, distinguished by what is held constant:

• Coffee-cup calorimeter (constant pressure): an open, insulated Styrofoam cup. It measures q at constant (atmospheric) pressure, so it directly measures ΔH (enthalpy change).
• Bomb calorimeter (constant volume): a sealed steel vessel. It measures q at constant volume, so it measures ΔE (internal energy change) directly.

The logic is always the same: the surroundings (the water/calorimeter) absorb whatever the system (the reaction or hot object) releases, and vice versa.

Worked Example

A 50.0 g sample of water is heated from 20.0 °C to 80.0 °C. How much heat was absorbed? (c_water = 4.184 J/(g·°C))

Step 1 — Find ΔT. ΔT = T_final − T_initial = 80.0 °C − 20.0 °C = 60.0 °C.

Step 2 — Apply q = mcΔT:

q = (50.0 g)(4.184 J/(g·°C))(60.0 °C) = 12,552 J = 12.6 kJ

The water absorbed 12.6 kJ (q > 0). Units cancel: g and °C drop out, leaving joules.

Second example — finding specific heat. A 45.0 g piece of metal at 100.0 °C is dropped into 100.0 g of water at 22.0 °C in a coffee-cup calorimeter. The final temperature is 25.6 °C. Find the specific heat of the metal.

Step 1 — Heat gained by water: q_water = (100.0 g)(4.184 J/(g·°C))(25.6 − 22.0 °C) = 100.0 × 4.184 × 3.6 = 1506 J.

Step 2 — Heat lost by metal: q_metal = −q_water = −1506 J.

Step 3 — Solve for c_metal. ΔT_metal = 25.6 − 100.0 = −74.4 °C.

q_metal = m c_metal ΔT_metal → c_metal = q_metal / (m ΔT_metal) = (−1506 J) / [(45.0 g)(−74.4 °C)] = 0.450 J/(g·°C)

Both negative signs cancel, giving a positive specific heat — as required.

Common Confusions

• c vs. C. Lowercase c is per gram (J/g·°C); capital C is for the whole object (J/°C). Don't mix them up.
• Forgetting the sign of ΔT. ΔT = final − initial. A cooling object has ΔT < 0, so its q is negative. Keep the signs straight or you'll get a negative specific heat (impossible).
• Calories vs. joules. 1 nutritional Calorie = 1000 cal = 4184 J = 4.184 kJ. A food "Calorie" (capital C) is a kilocalorie.
• °C vs. K. Only changes in temperature are equal in the two scales. Never plug an absolute temperature in kelvin into a °C ΔT formula carelessly — use the change.
• Which side is the system? The heat you calculate with the water's temperature change is the heat of the surroundings; the reaction's heat has the opposite sign.