Warming something costs energy, and every material charges its own rate. The relation is $Q = mc\Delta T$ — read aloud Q equals m c delta T. $Q$ is the heat added or removed, in kilojoules. $m$ is the mass being heated, in kilograms. $c$ is the specific heat capacity, in kJ/(kg·K) — the energy one kilogram needs for one kelvin of rise, and a property of the material, not of your process. And $\Delta T$ is the temperature change in kelvin, never a temperature the substance sits at. That distinction is worth a mark on every paper you will ever sit.

This is sensible heat: the kind a thermometer can sense. Put it in and the reading climbs; take it out and the reading falls. Water's $c$ is about 4.2 kJ/(kg·K), which is enormous — a third more than glycol, eight times steel — and that single number is why the world moves heat around buildings and plants in water rather than in anything else.

The relation solves four ways, and a good exam will ask for all of them: $m = \dfrac{Q}{c\,\Delta T}$ when a heat meter tells you how much you warmed, $c = \dfrac{Q}{m\,\Delta T}$ when a calorimeter is identifying a fluid, and $\Delta T = \dfrac{Q}{mc}$ when you know the duty and want the rise. Run the units through each one: kJ divided by (kJ/(kg·K) × K) leaves kilograms, and every term that belongs on the bottom is on the bottom.