8) Which one of the following objects, all initially at the same temperature, will be most efficient in losing heat?
a dull black box in vacuum
a polished silver box in air
a dull black box in air
a polished silver box in vacuum
a dull
To determine which object will be most efficient in losing heat, we need to consider the factors that affect heat transfer. There are three modes of heat transfer: conduction, convection, and radiation.
1. Conduction: This is the transfer of heat through direct contact between objects. In this case, all the objects are initially at the same temperature and are assumed to have the same composition. Therefore, conduction will not play a major role in heat transfer.
2. Convection: This is the transfer of heat through the movement of fluids (such as air or water). In this case, the only object in air is the polished silver box. Air is a better conductor of heat compared to a vacuum. Therefore, the polished silver box in air will experience more significant convective heat transfer.
3. Radiation: This is the transfer of heat through electromagnetic waves. All objects emit and absorb thermal radiation. The amount of thermal radiation emitted or absorbed depends on the material's emissivity and the temperature difference between objects. Dull black surfaces have higher emissivity than polished silver surfaces. Emissivity refers to the ability of a surface to emit and absorb thermal radiation. Therefore, the dull black surfaces (both in vacuum and air) will have higher radiant heat transfer.
Considering both convection and radiation, the dull black box in a vacuum will have the highest overall heat transfer efficiency. Since it has high emissivity for radiation and no convective heat transfer, the heat escaping from the box will be primarily through radiation.
Therefore, the dull black box in a vacuum will be the most efficient in losing heat among the given options.