Specific Heat And Thermal Conductivity

Finding $\Delta T$ for iron: Equation used: $$\Delta T=\frac{Q}{mc}$$ Given: $$Q=9000\text{ J}, m=2.0\text{ kg}, c=450\text{ J/(kg}\cdot\text{°C)}$$ Working: $$\Delta T=\frac{9000}{2.0\times 450}=\frac{9000}{900}$$ $$\Delta {T}_{\text{iron}}=10.0\text{ °C}$$ Finding $\Delta T$ for water: $$\Delta T=\frac{9000}{2.0\times 4186}=\frac{9000}{8372}$$ $$\Delta {T}_{\text{water}}=1.08\text{ °C}$$ Iron experiences the greater temperature change. Specific heat is the energy required per kilogram per degree. Iron’s specific heat is much lower than water’s, so each joule of energy produces a larger temperature change in iron.

Doubling the mass does not change the specific heat. Specific heat is an intrinsic property that depends on the arrangement and interactions of the material’s atoms, not on the amount of material present. A larger sample requires more total energy $Q$ for the same $\Delta T$, but the ratio $\frac{Q}{m\Delta T}$ remains constant.

Derivation: Equilibrium Temperature Starting point: Conservation of energy — in an isolated system, the total energy is constant. The energy lost by the hot object equals the energy gained by the cool object. $${Q}_{\text{lost}}+{Q}_{\text{gained}}=0$$ Step 1 — Substitute the expression $Q=mc\Delta T$ for each object: $${m}_{1}{c}_{1}({T}_{f}-{T}_{1i})+{m}_{2}{c}_{2}({T}_{f}-{T}_{2i})=0$$ Step 2 — Expand both terms: $${m}_{1}{c}_{1}{T}_{f}-{m}_{1}{c}_{1}{T}_{1i}+{m}_{2}{c}_{2}{T}_{f}-{m}_{2}{c}_{2}{T}_{2i}=0$$ Step 3 — Collect all terms containing ${T}_{f}$ on one side: $${T}_{f}({m}_{1}{c}_{1}+{m}_{2}{c}_{2})={m}_{1}{c}_{1}{T}_{1i}+{m}_{2}{c}_{2}{T}_{2i}$$ Step 4 — Isolate ${T}_{f}$: $${T}_{f}=\frac{{m}_{1}{c}_{1}{T}_{1i}+{m}_{2}{c}_{2}{T}_{2i}}{{m}_{1}{c}_{1}+{m}_{2}{c}_{2}}$$ Result: $${T}_{f}=\frac{{m}_{1}{c}_{1}{T}_{1i}+{m}_{2}{c}_{2}{T}_{2i}}{{m}_{1}{c}_{1}+{m}_{2}{c}_{2}}$$ The final temperature is a weighted average of the initial temperatures, weighted by each object’s thermal mass ($mc$).

Because both rods share the same $A$, $L$, and $\Delta T$, the ratio depends only on thermal conductivity. Equation used: $$\frac{(Q/\Delta t{)}_{\text{Cu}}}{(Q/\Delta t{)}_{\text{glass}}}=\frac{{k}_{\text{Cu}}}{{k}_{\text{glass}}}$$ Working: $$\frac{(Q/\Delta t{)}_{\text{Cu}}}{(Q/\Delta t{)}_{\text{glass}}}=\frac{400}{0.80}=500$$ Energy transfers through the copper rod 500 times faster than through the glass rod.

The rate of energy transfer is halved. The rate is inversely proportional to thickness $L$, because $\frac{Q}{\Delta t}=\frac{kA\Delta T}{L}$. Doubling $L$ doubles the denominator, so the rate decreases to $\frac{1}{2}$ of its original value.

The expression is consistent with the claim. If $A$ or $L$ changes, the rate of energy transfer $\frac{Q}{\Delta t}$ changes, but $k$ remains constant because it can be isolated as $k=\frac{QL}{\Delta t\cdot A\cdot \Delta T}$. The dimensions appear as separate variables, so changing them does not alter $k$. This confirms that $k$ depends only on the material’s atomic arrangement and interactions, not on the sample’s size or shape.

Cut the unknown material into a slab of known cross-sectional area $A$ and thickness $L$, measured with a ruler. Attach one face to a heat source at a known steady temperature and the other face to a heat sink at a lower steady temperature. Measure both surface temperatures with thermometers to determine $\Delta T$. Measure the energy transferred through the slab over a measured time interval $\Delta t$ using a calorimeter on the cool side. Calculate $k$ from $k=\frac{QL}{\Delta t\cdot A\cdot \Delta T}$. To reduce uncertainty, repeat the measurement for several values of $\Delta T$ and plot $\frac{Q}{\Delta t}$ on the vertical axis against $\Delta T$ on the horizontal axis. The slope of the best-fit line equals $\frac{kA}{L}$, from which $k$ can be extracted.