$Q = mc\,\Delta T$ — Q equals m c delta-T — is one fact wearing four outfits: solve it for $Q$, for $m$, for $c$ or for $\Delta T$, and it answers four different exam questions. Push the units through before any arithmetic: $\mathrm{kg} \times \mathrm{J/(kg\cdot{}^{\circ}C)} \times {}^{\circ}\mathrm{C} = \mathrm{J}$ — everything cancels to joules, so the machine is honest. If your rearrangement's units refuse to cancel, it is wrong — no appeal. (Units that do work out never prove it right.)

Two flags for this lesson. First: $\Delta T$ is a difference, and a difference of 30 C° is a difference of 30 K — for $\Delta T$, Celsius is safe as houses. Second: $c$ arrives per kilogram, and problems love to hand you grams. Convert before the socket — the g-vs-kg slip is worth a thousand-fold error, and this lesson sets that trap on purpose.