14.3 - Specific heat capacity

Energy for a temperature rise

Why do equal masses of water and aluminium not warm by the same amount when they receive the same energy? The materials require different amounts of energy for each kilogram to rise by one kelvin. That difference is measured by their specific heat capacities.

The word specific means per unit mass. A large sample needs more energy than a small sample of the same material for the same temperature rise. Heat capacity, without “specific”, refers to the whole object's energy requirement per kelvin and has unit $\mathrm{J\,K^{-1}}$. Specific heat capacity has unit $\mathrm{J\,kg^{-1}\,K^{-1}}$.

For fixed $Q$ and $m$, a larger $c$ gives a smaller rise. Using representative values of $4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$ for water and $900\,\mathrm{J\,kg^{-1}\,K^{-1}}$ for aluminium, each $0.50\,\mathrm{kg}$ sample receiving $4500\,\mathrm{J}$ would rise by about $2.1\,\mathrm{K}$ and $10\,\mathrm{K}$, respectively. This is why energy input alone does not determine temperature change.

Heating is an energy transfer, not a quantity called heat stored in the sample. A temperature rise is evidence of a change in the material's state; internal energy is a property of the material. In the next lesson you will see a case where energy is transferred while temperature stays constant during a change of state.

State what the definition means in a physical situation.

Heating calculations and energy balances

Begin with the system and the energy transfers, then choose the equation. A mass in grams must become kilograms if $c$ is in $\mathrm{J\,kg^{-1}\,K^{-1}}$. A cooling object loses energy; it is often simplest to calculate the positive amount lost from the positive temperature drop and state its direction separately.

When a hot object enters cooler water, write energy lost by the hot object equals energy gained by the cooler parts, provided there is negligible transfer to the surroundings. Include the container if its thermal response is significant. You do not obtain the common temperature by averaging the starting temperatures unless the two total heat capacities are equal.

For a constant useful heating power $P$ acting for time $t$, the transferred energy is $Pt$. The power relation is not on the data sheet: recall power as energy transferred per unit time. Combining it with $Q=mc\Delta T$ gives $Pt=mc\Delta T$, with $P$ in watts and $t$ in seconds. It applies only when $P$ is the power actually heating the stated sample.