Q.Explain why:

1. Key idea: Kirchhoff’s law — at thermal equilibrium, emissivity equals absorptivity ($\varepsilon = \alpha$). A body with large reflectivity has small absorptivity ($\alpha \approx 0$), hence small emissivity ($\varepsilon \approx 0$). A poor absorber is a poor emitter.
2. Key idea: Thermal conductivity and specific heat capacity. Brass has much higher thermal conductivity than wood. On a chilly day, the brass tumbler rapidly conducts heat away from your hand, making it feel colder. Wood conducts heat slowly, so your skin’s heat is not drawn away as fast.
3. Key idea: Optical pyrometers measure brightness temperature, assuming the source is a blackbody ($\varepsilon = 1$). A red-hot iron piece in the open has emissivity $\varepsilon < 1$ (it reflects some light), so it appears less bright than a blackbody at the same true temperature. The pyrometer, interpreting lower brightness as lower temperature, gives a value too low. Inside the furnace, the iron piece is surrounded by walls at the same temperature — it is in thermal equilibrium with its environment, so its effective emissivity approaches 1 (cavity radiation), and the pyrometer reads correctly.
4. Key idea: The greenhouse effect. Earth’s atmosphere traps infrared radiation emitted by the surface, reducing net heat loss to space. Without this blanket, the Earth would radiate its heat directly into space, leading to a much lower average surface temperature (about $-18^\circ$C instead of $+15^\circ$C). …

(a) A body with large reflectivity is a poor emitter

Concept: Kirchhoff’s law of thermal radiation states that at a given temperature and wavelength, the emissivity $\epsilon$ of a surface equals its absorptivity $\alpha$:

A body that reflects most incident radiation (high reflectivity $r$) must absorb very little, because for an opaque surface $r + \alpha = 1$. So $\alpha$ is small, and by Kirchhoff’s law, $\epsilon$ is also small. A poor absorber is necessarily a poor emitter.

1. Reflectivity and absorptivity are complementary. For an opaque body, all incident radiation that is not reflected is absorbed: $r + \alpha = 1$. High $r$ means low $\alpha$.
2. Kirchhoff’s law ties emission to absorption. At thermal equilibrium, the rate at which a body emits radiation equals the rate at which it would absorb radiation from a blackbody at the same temperature. Hence $\epsilon = \alpha$.
3. Conclusion. If $\alpha$ is small, $\epsilon$ is small — the body radiates poorly. A shiny, highly reflective surface (like polished silver) is a terrible emitter of thermal radiation.

(b) A brass tumbler feels much colder than a wooden tray on a chilly day

Concept: The sensation of “cold” or “hot” depends on the rate of heat flow between your skin and the object, not on the object’s temperature alone. Both the brass tumbler and the wooden tray are at the same ambient temperature (say, $10^\circ$C), but brass has a much higher thermal conductivity than wood.

1. Thermal conductivity difference. Brass has a thermal conductivity $k \approx 110\ \text{W m}^{-1}\text{K}^{-1}$, while wood has $k \approx 0.1\ \text{W m}^{-1}\text{K}^{-1}$ — a factor of over 1000.
2. Heat transfer from your hand. When you touch the brass, heat flows rapidly from your warmer skin ($\approx 33^\circ$C) into the cold brass. The high $k$ means a large heat current per unit area:

where $d$ is the depth of penetration. Your skin’s temperature drops quickly, and nerve endings signal “cold.”

3. Wood feels less cold. Wood conducts heat so slowly that the heat from your hand barely penetrates. The surface of the wood warms up locally, reducing the temperature gradient. The heat loss from your skin is much smaller, so it feels less cold — even though both objects are at the same temperature.

(c) Optical pyrometer gives too low a value for red-hot iron in the open, but correct value when the same piece is in a furnace

Concept: An optical pyrometer measures temperature by comparing the brightness of the object at a specific wavelength (usually red) to that of a calibrated filament. It is calibrated assuming the object is a blackbody (emissivity $\epsilon = 1$). Real objects have $\epsilon < 1$, so they emit less radiation at a given temperature than a blackbody would.

1. In the open. A red-hot iron piece in open air has an emissivity $\epsilon < 1$ (typically $\approx 0.3$–$0.5$ for oxidised iron). The pyrometer sees a lower intensity than a blackbody at the same temperature would emit. Since the pyrometer interprets intensity as temperature using the blackbody calibration, it underestimates the true temperature.
2. Inside a furnace. When the iron piece is inside a furnace, the furnace walls are also hot. The piece is surrounded by a cavity at nearly the same temperature. Multiple reflections inside the cavity make the effective emissivity approach 1 — the cavity behaves like a blackbody. The pyrometer now sees radiation that is essentially blackbody radiation, so the reading is correct.
3. Why the cavity works. A small hole in a furnace wall is a classic blackbody: any radiation entering the hole is trapped by multiple reflections and almost completely absorbed. Conversely, radiation emerging from the hole is characteristic of a blackbody at the furnace temperature. The iron piece inside the furnace is part of this cavity, so the radiation it emits (plus reflected radiation from the walls) mimics a blackbody.

(d) The earth without its atmosphere would be inhospitably cold

Concept: The Earth’s atmosphere acts as a greenhouse blanket — it is largely transparent to incoming solar radiation (visible light) but absorbs and re-emits a large fraction of the outgoing infrared radiation from the Earth’s surface.

1. Energy balance with atmosphere. Incoming solar radiation (shortwave) heats the surface. The surface radiates infrared (longwave) back. Greenhouse gases ($\text{CO}_2$, $\text{H}_2\text{O}$, $\text{CH}_4$) absorb much of this outgoing IR and re-radiate it in all directions, including back down to the surface. This downward infrared radiation keeps the surface temperature about $33^\circ$C warmer than it would be without an atmosphere.
2. Without atmosphere. There would be no greenhouse gases to trap outgoing IR. The Earth’s surface would radiate directly into space. The effective radiating temperature of the Earth (the temperature at which it balances solar input) is about $255\ \text{K}$ ($-18^\circ$C). With an atmosphere, the actual average surface temperature is about $288\ \text{K}$ ($15^\circ$C). Without it, the surface would be frozen — inhospitably cold.
3. Additional effects. The atmosphere also distributes heat via convection and weather patterns, but the dominant effect is the radiative greenhouse effect. Without it, even the equator would be much colder at night, and the planet would be largely ice-covered. …

Key insight — one unifying idea threads all five parts: heat transfer follows the path of least resistance, and radiation specifically obeys Kirchhoff's law ($\epsilon=\alpha$).

1. Think of a mirror: it reflects nearly all incoming light, so it absorbs almost none — and whatever a surface is bad at absorbing, Kirchhoff's law says it's equally bad at emitting.
2. The 'cold' sensation isn't about an object's actual temperature but about how fast it can pull heat out of your hand. Brass's thermal conductivity is roughly 1000x wood's, so it drains warmth from your skin far faster, which your nerves read as 'cold' even though both are at the same temperature.
3. A pyrometer really compares brightness to an ideal ($\epsilon=1$) reference. The open iron is a 'leaky' emitter ($\epsilon<1$), so it looks dimmer — and hence cooler — than it truly is; inside a furnace cavity, radiation bounces around and gets re-absorbed so many times that the effective emissivity is forced toward 1, and the same physical temperature now reads correctly. …