Name the heat-transfer mechanism that can occur through a vacuum.

A slab has \(k=0.20\,\mathrm{W\,m^{-1}\,K^{-1}}\), area \(1.5\,\mathrm{m^2}\), thickness \(0.050\,\mathrm{m}\), and temperature difference \(10\,\mathrm{K}\). Find the conduction heat current.

If a conducting wall's thickness is doubled while \(k\), \(A\), and \(\Delta T\) are unchanged, what happens to the heat current?

A window has thermal resistance \(R=0.80\,\mathrm{K\,W^{-1}}\) and a temperature difference of \(16\,\mathrm{K}\). Find the heat current through it.

A surface of area \(0.30\,\mathrm{m^2}\) has convection coefficient \(h_c=12\,\mathrm{W\,m^{-2}\,K^{-1}}\). The surface is \(8.0\,\mathrm{K}\) hotter than the surrounding fluid. Estimate the convective heat current.

A black surface with \(e=0.95\), area \(0.10\,\mathrm{m^2}\), and temperature \(350\,\mathrm{K}\) radiates into surroundings at \(300\,\mathrm{K}\). Find the net radiative heat loss using \(\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}\).

Why does a metal handle feel colder than a wooden handle at the same room temperature?

Two slabs are in series. Slab 1 has \(R_1=0.40\,\mathrm{K\,W^{-1}}\), slab 2 has \(R_2=0.90\,\mathrm{K\,W^{-1}}\), and the total temperature difference is \(26\,\mathrm{K}\). Find the steady heat current and the temperature drop across slab 2.

A wall has two parallel heat-flow paths. Path A has \(R_A=2.0\,\mathrm{K\,W^{-1}}\), path B has \(R_B=5.0\,\mathrm{K\,W^{-1}}\), and both see the same \(20\,\mathrm{K}\) temperature difference. Find the total heat current.

A surface radiates net power \(H\) at temperature \(T\) into very cold surroundings. If its absolute temperature is doubled and \(A,e\) are unchanged, by what factor does \(H\) change? Explain why Celsius cannot be used for this ratio.

Derive the effective thermal resistance for two conducting slabs in series, with thicknesses \(L_1,L_2\), conductivities \(k_1,k_2\), common area \(A\), and steady one-dimensional heat flow.

A hot object loses heat by radiation \(e\sigma A(T^4-T_s^4)\) and convection \(h_cA(T-T_s)\). For a small excess temperature \(\theta=T-T_s\), derive the approximate total heat-loss law linear in \(\theta\).