Answer
See answers.
Work Step by Step
a. The more power radiated, the better the efficiency of the bulb. Use Equation 14–6. Let “H” stand for the halogen bulb and “I” for the incandescent bulb.
$$\frac{P_H-P_I}{P_I}=\frac{T_H^4-T_I^4}{T_I^4}$$
$$=\frac{(2900K)^4-(2700K)^4}{(2700K)^4}=0.331\approx 33\%$$
The halogen bulb is about 33% more efficient than the traditional incandescent bulb.
b. We just determined that the halogen bulb is 33% more efficient than the incandescent bulb. In other words, the old 100W power for the old bulb can be decreased by a factor of 1.33, and the halogen will give out roughly the same amount of radiation.
$$P_H=\frac{100W}{1.33}=75W$$
