How far the temperature moves, and how fast it gets there

Specific heat $c$ is the energy needed per kilogram per degree, so it appears in $Q = mc\,\Delta T$. Solve for the thing that actually responds:

$$\Delta T = \frac{Q}{mc}.$$

$c$ sits in the denominator. A material with a large $c$ therefore resists changing temperature: the same energy moves it less far. Water's $c \approx 4186$ J/(kg$\cdot$K) is about five times aluminum's and about ten times copper's, so a joule delivered to water buys about a tenth the temperature rise it buys in copper.

Reading the big number as a big response gets the physics backward and takes a set of familiar facts with it. Water is slow to heat and slow to cool. The sea beside a beach stays milder than the sand all day and all night. A car radiator uses water because it can absorb a lot of energy without its own temperature running away.

Put two substances in an insulated container. Whatever energy one loses, the other gains:

$$m_1c_1(T_f - T_1) + m_2c_2(T_f - T_2) = 0.$$

Solve that and $T_f$ is the $mc$-weighted average of the two starting temperatures. It lands nearer the side with the larger $mc$, because that side has more thermal inertia. The midpoint is correct only in the one case where both sides have the same $mc$.

Mix $100$ g of water at $80^\circ$C with $400$ g of water at $20^\circ$C. The specific heats cancel, the cold side has four times the mass, and

$$T_f = \frac{100(80) + 400(20)}{500} = 32^\circ\text{C},$$

not $50^\circ$C. Two checks catch most errors here: the answer must land between the two starting temperatures, and one side's loss must equal the other's gain.

For a slab of area $A$ and thickness $L$ with a temperature difference $\Delta T$ across it,

$$\frac{Q}{\Delta t} = \frac{kA\,\Delta T}{L}, \qquad \text{measured in watts}.$$

Read the dependences off it directly. The rate rises with conductivity, with area, and with the temperature difference, and it falls with thickness, so doubling the insulation halves the rate. To get an amount of energy you multiply the rate by a time; quoting a wattage as though it were joules is the same error as quoting a speed as a distance.

Nothing in that expression is a temperature of the material. A metal bench and a wooden bench in one room are at the same temperature, and the metal feels colder because its large $k$ pulls energy out of your hand quickly. Your skin reports a rate; you interpret it as a state. That is the misreading, and it is why the same two benches feel equally warm on a hot day, with the metal now the more uncomfortable one.

One boundary case is worth stating because it breaks the $Q = mc\,\Delta T$ habit. While a substance is melting or boiling, energy keeps arriving and the temperature does not move: the energy goes into separating the particles rather than speeding them up.

So an ice-water mixture sits at $0^\circ$C for as long as any ice remains, however hard you heat it, and a pot of boiling water stays at $100^\circ$C whether the burner is on low or high. Turning the burner up makes it boil away faster, not hotter.

Any problem that carries a substance through melting or boiling needs the run split into stages, with $mc\,\Delta T$ used on the sloping parts and the phase-change energy accounted for separately on the flat parts.