Vardaan Institute - Detailed Solutions: Master Sheet (Thermal Properties of Matter)

Section A: Thermal Expansion Solutions

Show that $\gamma = 3\alpha$.

Sol: $V' = L'^3 = (L_0(1+\alpha\Delta T))^3 = V_0(1 + 3\alpha\Delta T + \dots) \implies \gamma = 3\alpha$.

Aluminium sphere calorimetry calculation.

Sol: Heat lost $= m_{al} s_{al} (100 - 23) = 0.047 \times s_{al} \times 77 = 3.619 s_{al}$. Heat gained $= m_w s_w (23 - 20) = 0.25 \times 4186 \times 3 = 3139.5\text{ J}$. $s_{al} = \frac{3139.5}{3.619} \approx 867\text{ J/kg}\cdot\text{K}$.

Section B: Conduction & Cooling Solutions

Calculate heat transfer rate $\frac{dQ}{dt} = K A \frac{T_1 - T_2}{L}$.

Sol: $\frac{dQ}{dt} = 200 \times (10 \times 10^{-4}) \times \frac{100 - 0}{0.5} = 0.2 \times 200 = 40\text{ W}$.

Newton's Law of Cooling problem.

Sol: Case 1: $\frac{80 - 50}{5} = k \left(\frac{80+50}{2} - 20\right) \implies 6 = k(45) \implies k = \frac{2}{15}$. Case 2: $\frac{50 - 30}{t} = k \left(\frac{50+30}{2} - 20\right) \implies \frac{20}{t} = \frac{2}{15}(20) \implies t = 15\text{ minutes}$.

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