Radiation of Heat

Stefan-Boltzmann Law and Net Radiation Heat Transfer

Conditions: Black body in an environment at constant absolute temperature

• The total radiant energy emitted per second per unit area by a black body is directly proportional to the fourth power of its absolute temperature T: $E = \sigma T^{4}$
• If a black body of surface area A at temperature T is placed in an enclosure at temperature T₀, it emits energy at a rate Q_emit = \sigma A T^{4}
• Simultaneously, the body absorbs radiant energy from the enclosure at a rate Q_abs = \sigma A T_{0}^{4}
• The net rate of heat loss by radiation is given by subtracting absorption from emission: $Q_{net} = \sigma A (T^{4} - T_{0}^{4})$

Result: $Q_{net} = \sigma A (T^{4} - T_{0}^{4})$

Newton's Law of Cooling from Stefan-Boltzmann Law

Conditions: Small temperature difference between the body and surroundings ((T - T_{0}) \ll T_{0})

• Start with the net rate of heat loss from Stefan-Boltzmann law: - \frac{dQ}{dt} = \sigma A e (T^{4} - T_{0}^{4}), where e is emissivity.
• Factor the temperature term: T^{4} - T_{0}^{4} = (T^{2} - T_{0}^{2})(T + T_{0}) = (T - T_{0})(T + T_{0})(T^{2} + T_{0}^{2}).
• Substitute T \approx T_{0} for the terms other than (T - T_{0}): T^{4} - T_{0}^{4} \approx (T - T_{0})(2T_{0})(2T_{0}^{2}) = 4T_{0}^{3}(T - T_{0}).
• Substitute this approximation back into the rate equation and write dQ = mc dT to obtain \frac{dT}{dt} = - \frac{4\sigma A e T_{0}^{3}}{mc}(T - T_{0}) = - K(T - T_{0}).

Result: \frac{dT}{dt} = - K(T - T_{0})

Blackbody Spectral Radiance (Planck's Distribution Curves)

Draw: Curves of spectral emissive power E_{\lambda} versus wavelength \lambda for various absolute temperatures T, showing asymmetric peaks that shift to shorter wavelengths as temperature increases.

Interpret: The area under each curve represents total emissive power \sigma T^{4}. The peak shifts toward shorter wavelengths with rising temperature, illustrating Wien's displacement law.

Exam use: Directly tested in questions regarding shifts in peak wavelength and verification of Stefan-Boltzmann law integration.

Newton's Law of Cooling Experimental Verification Curve

Draw: A semi-log plot of ln(T - T_{0}) versus time t, yielding a straight line with a negative slope equal to -K.

Interpret: Linearity of the semi-log plot confirms the exponential decay nature of Newton's law of cooling.

Exam use: Interpreting experimental gradient to find cooling constant K for a calorimeter.

Ferry's Black Body Apparatus

Draw: A schematic double-walled hollow sphere with a small aperture and an internal conical projection opposite the aperture.

Interpret: Any radiation entering through the aperture undergoes multiple internal reflections, with a negligible fraction escaping, achieving an effective absorptivity close to 1.

Exam use: Conceptual questions on the practical realization of a perfectly black body.

A metal sphere of radius 0.1 m is heated to 500 K and placed in an environment at 300 K. Assuming it behaves as a black body, calculate the net rate of heat loss by radiation. (Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} W m^{-2} K^{-4})$

• Determine the surface area of the sphere: $A = 4 \pi r^{2} = 4 \times \pi \times (0.1)^{2} = 0.04 \pi \approx 0.12566 m^{2}$.
• Apply the net radiation formula: $Q_{net} = \sigma A (T^{4} - T_{0}^{4})$.
• Calculate T^{4} - T_{0}^{4} = 500^{4} - 300^{4} = 6.25 \times 10^{10} - 0.81 \times 10^{10} = 5.44 \times 10^{10} K^{4}.
• Substitute values: $Q_{net} = (5.67 \times 10^{-8}) \times (0.12566) \times (5.44 \times 10^{10}) = 387.6 W$.

Answer: 387.6 W

The spectral emissive power of a black body peaks at a wavelength of \lambd$a_{m} = 500 nm at a certain temperature T_{1}$. If the temperature is increased such that the peak wavelength shifts to 400 nm, find the percentage change in the total emissive power of the black body.

• Use Wien's displacement law to find the new temperature T_{2}: \lambda_{1} $T_{1} = \lambda_{2} T_{2} \implies 500 T_{1} = 400 T_{2} \implies T_{2} = \frac{5}{4} T_{1} = 1.25 T_{1}$.
• The total emissive power is proportional to T^{4}: $E_{1} = \sigma T_{1}^{4}$ and $E_{2} = \sigma T_{2}^{4} = \sigma (1.25 T_{1})^{4} = 2.4414 \sigma T_{1}^{4}$.
• Calculate the percentage change: \frac{E_{2} - E_{1}}{E_{1}} \times $100 = (2.4414 - 1) \times 100 = 144.14%$.

Answer: 144.14% increase

A body cools from 80°C to 64°C in 5 minutes in a room maintained at 24°C. Find the time taken by the body to cool from 64°C to 52°C, assuming Newton's law of cooling holds.

• Find the average temperature for the first interval: $T_{avg1} = \frac{80 + 64}{2} = 72°C$.
• Apply Newton's law of cooling differential form: \frac{T_{1} - T_{2}}{t} = K(T_{avg} - T_{0}) \implies \frac{80 - 64}{5} = K(72 - 24) \implies \frac{16}{5} = K(48) \implies $K = \frac{16}{5 \times 48} = \frac{1}{15} min^{-1}$.
• Find the average temperature for the second interval: $T_{avg2} = \frac{64 + 52}{2} = 58°C$.
• Apply the equation for the second interval: \frac{64 - 52}{t'} = K(58 - 24) \implies \frac{12}{t'} = \frac{1}{15} (34) \implies t' = \frac{12 \times 15}{34} = \frac{180}{34} \approx 5.29 min.

Answer: 5.29 minutes