convection\nsample problem 1\n1. a photographic enlarger has as its light source a 25 - w lamp in a metal…

convection\nsample problem 1\n1. a photographic enlarger has as its light source a 25 - w lamp in a metal hood that keeps light from escaping except through enlarger lens. when the dark room temperature is 20°c , the metal hood is at a temperature of 50°c. if the 25 - w lamp is replaced by a 50 - w lamp, what is the new temperature of the hood?

convection\nsample problem 1\n1. a photographic enlarger has as its light source a 25 - w lamp in a metal hood that keeps light from escaping except through enlarger lens. when the dark room temperature is 20°c , the metal hood is at a temperature of 50°c. if the 25 - w lamp is replaced by a 50 - w lamp, what is the new temperature of the hood?

Answer

Explanation:

Step1: Assume power - temperature relationship

Assume that the power dissipated by the lamp is proportional to the temperature difference between the hood and the room. Let $P_1$ be the initial power, $T_1$ be the initial hood temperature, $T_{room}$ be the room temperature, $P_2$ be the new power, and $T_2$ be the new hood temperature. The relationship can be written as $\frac{P_1}{T_1 - T_{room}}=\frac{P_2}{T_2 - T_{room}}$.

Step2: Substitute given values

We know that $P_1 = 25\ W$, $T_1=50^{\circ}C$, $T_{room}=20^{\circ}C$, and $P_2 = 50\ W$. Substituting these values into the equation $\frac{25}{50 - 20}=\frac{50}{T_2 - 20}$.

Step3: Cross - multiply and solve for $T_2$

Cross - multiplying gives $25(T_2 - 20)=50\times30$. Expanding the left - hand side: $25T_2-500 = 1500$. Then, adding 500 to both sides: $25T_2=2000$. Dividing both sides by 25, we get $T_2 = 80^{\circ}C$.

Answer:

$80^{\circ}C$