Wien’s displacement law
Why does a hotter object appear blue or white, while a cooler object appears red?
Hotter objects emit more radiation at shorter wavelengths, which correspond to blue or white light, while cooler objects emit more radiation at longer wavelengths, corresponding to red light.
A hot plate has a surface area of 0.025m2 and a constant temperature of 150°C. Its surroundings are kept at a temperature of 20°C. We assume that the hot plate behaves like a black body and that thermal radiation is the only energy transfer mechanism. Determine the net power exchanged by the exposed face of the hot plate with the surroundings.
The absolute temperatures of the hot plate and the surroundings are Thot plate = 423 K and Tsurroundings = 293K. The net power is the difference between the power radiated by the hot plate and the power it absorbs from the surroundings: Pnet = σA(T4hot plate - T4suroundings) = 5.67 × 10-8 x 0.025(4234 - 2934) = 35 W
When you touch a warm metal spoon in a bowl of hot soup, which process transfers heat through the spoon, and which process heats the soup?
Conduction transfers heat through the metal spoon, and convection transfers heat through the soup.
A student uses a prism spectrophotometer to investigate how the intensity of light from two incandescent lamps A and B varies with wavelength. Given peak wavelength of lamp A is 1200nm and peak wavelength of lamp B is 1400nm, estimate the temperatures Ta and Tb of the light-emitting wire filaments of each lamp.
Ta = (2.9*10-3)/(1200*10-9) ≈ 2400K
Tb = (2.9*10-3)/(1400*10-9) ≈ 2100K
The Sun has a diameter of 1.4 × 109 m and a surface temperature of 5800K. Calculate:
a. the power radiated by one square metre of the surface of the Sun
b. the total energy radiated by the Sun in one day.
a. P=σT4 = 5.67 × 10-8 × 58004 = 6.4 × 107W
b. The surface area of the Sun is 4π((1.4x109)/2)=6.2 × 1018 m2
The energy radiated in one day is 6.4 × 107 × 6.2 × 1018 x 24 × 60 × 60 = 3.4 × 1031 J
A metal rod of length L = 0.5 m and cross-sectional area A = 0.02 m² connects two reservoirs, maintaining temperatures of T₁ = 373 K and T₂ = 273 K at its ends. The thermal conductivity of the metal is k = 50 W·m⁻¹·K⁻¹.
(a) Calculate the rate of thermal energy Q̇/t
Q/t = k * A * (T2 - T1) / x
Q/t = 50 * 0.02 * 100 / 0.5 = 200W