iGCSE Physics (0625) 2.3.3 Radiation-Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Detailed solution:
To cool the liquid quickly, the rate of thermal energy transfer to the surroundings must be maximized. Surface color significantly affects this; dull black surfaces are the best emitters of infrared radiation compared to shiny silver ones. Additionally, attaching copper fins increases the total surface area $A$ available for both radiation and convection. Since the rate of emission is directly proportional to the surface area, the design with fins provides a higher cooling rate. Therefore, the combination of a black surface for high emissivity and fins for increased area ensures the most rapid cooling.
▶️ Answer/Explanation
Detailed solution:
Thermal energy naturally flows from regions of higher temperature to lower temperature via conduction through solids.
Since the middle of the filament is at a very high temperature and the ends are cooler, energy is conducted toward the base.
Option B is incorrect because the rate of radiation depends on temperature; therefore, the hotter middle radiates more energy than the cooler ends.
Option C is incorrect as energy is also transferred via infrared radiation (heat) and conduction, not just visible light.
Option D is incorrect because power is proportional to the square of the voltage, $P = \frac{V^{2}}{R}$.
If the potential difference $V$ is halved to $\frac{V}{2}$, the power output becomes $\frac{1}{4}$ of its original value, assuming resistance $R$ is constant.
▶️ Answer/Explanation
Detailed solution:
The total power $P$ emitted by a star depends on its surface area $A$ and its surface temperature $T$.
Since star $X$ has a larger diameter than star $Y$, it has a significantly larger surface area ($A_X > A_Y$).
The problem states both stars emit the same total power, so $P_X = P_Y$.
For a larger object to emit the same total energy as a smaller one, its emission per unit area must be lower.
Since the rate of emission increases with temperature, star $X$ must be cooler to compensate for its larger size.
Therefore, the surface temperature of $X$ is less than the surface temperature of $Y$.