Question

convection
sample problem 1

1. 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?

Explanation:

Step1: Assume power - heat relationship

Assume the power dissipated by the lamp is proportional to the heat generated, and the heat transfer from the hood to the surroundings is by convection. The rate of heat transfer $Q$ by convection is given by $Q = hA\Delta T$, where $h$ is the convective heat - transfer coefficient, $A$ is the surface area of the hood, and $\Delta T$ is the temperature difference between the hood and the surroundings. At steady - state, the power input from the lamp is equal to the power output (heat transfer to the surroundings). So, $P = hA\Delta T$.

Let the initial power be $P_1 = 25\ W$, the initial temperature of the hood be $T_1=50^{\circ}C$, and the room temperature be $T_{room}=20^{\circ}C$. Then $P_1 = hA(T_1 - T_{room})$.

Let the new power be $P_2 = 50\ W$, and the new temperature of the hood be $T_2$. Then $P_2 = hA(T_2 - T_{room})$.

Step2: Set up ratio of power equations

Dividing the second equation by the first equation:
$$\frac{P_2}{P_1}=\frac{hA(T_2 - T_{room})}{hA(T_1 - T_{room})}$$
Since $h$ and $A$ are constant, they cancel out. We know that $P_1 = 25\ W$, $P_2 = 50\ W$, $T_1 = 50^{\circ}C$, and $T_{room}=20^{\circ}C$. Substituting the values:
$$\frac{50}{25}=\frac{T_2 - 20}{50 - 20}$$
$$2=\frac{T_2 - 20}{30}$$

Step3: Solve for $T_2$

Cross - multiply:
$$2\times30=T_2 - 20$$
$$60=T_2 - 20$$
Add 20 to both sides:
$$T_2=60 + 20=80^{\circ}C$$

Answer:

$80^{\circ}C$