Solution
Step 1: Define the heat transfer mechanism
Let the resistance of a single cylindrical resistor be $R$. The problem states that the curved surfaces are thermally insulated. Heat transfer occurs only through the circular ends (cross-sectional areas).
According to Newton’s Law of Cooling (or the given proportionality), the rate of heat loss $H$ is proportional to the temperature difference between the object and the surroundings:
$$H = k(\theta – \theta_0)$$Where:
- $k$ is a cooling constant proportional to the exposed surface area.
- $\theta$ is the steady-state temperature of the resistor.
- $\theta_0$ is the room temperature ($20^\circ\text{C}$).
Step 2: Analysis of the first case (Single Resistor)
When a single resistor is connected to voltage $V$, the power dissipated is:
$$ P_1 = \frac{V^2}{R} $$At steady state, the rate of heat generation equals the rate of heat loss. Since heat is only lost from the two circular ends, let the cooling constant be $k$.
$$ \frac{V^2}{R} = k(\theta_1 – \theta_0) \quad \dots \text{(Equation 1)} $$Step 3: Analysis of the second case (Three Resistors in Series)
When three identical resistors are connected in series, the total resistance becomes:
$$ R_{\text{total}} = R + R + R = 3R $$The power dissipated in this combination is:
$$ P_2 = \frac{V^2}{3R} $$Key Insight: The resistors are connected in series (end-to-end) and their curved surfaces are insulated. Therefore, heat can only escape from the two extreme circular ends of the “pack.” The internal circular faces are in contact with each other and do not transfer heat to the surroundings. Consequently, the surface area available for cooling in the pack is identical to that of the single resistor.
Thus, the cooling constant $k$ remains the same. For the steady-state temperature $\theta_2$:
$$ \frac{V^2}{3R} = k(\theta_2 – \theta_0) \quad \dots \text{(Equation 2)} $$Step 4: Solve for $\theta_2$
Dividing Equation 1 by Equation 2:
$$ \frac{ \left( \frac{V^2}{R} \right) }{ \left( \frac{V^2}{3R} \right) } = \frac{k(\theta_1 – \theta_0)}{k(\theta_2 – \theta_0)} $$ $$ 3 = \frac{\theta_1 – \theta_0}{\theta_2 – \theta_0} $$Rearranging to solve for $\theta_2$:
$$ 3(\theta_2 – \theta_0) = \theta_1 – \theta_0 $$ $$ 3\theta_2 – 3\theta_0 = \theta_1 – \theta_0 $$ $$ 3\theta_2 = \theta_1 + 2\theta_0 $$ $$ \theta_2 = \frac{2\theta_0 + \theta_1}{3} $$Step 5: Substitution
Given values:
- $\theta_0 = 20^\circ\text{C}$ (Room temperature)
- $\theta_1 = 38^\circ\text{C}$
Substituting these values:
$$ \theta_2 = \frac{2(20) + 38}{3} $$ $$ \theta_2 = \frac{40 + 38}{3} $$ $$ \theta_2 = \frac{78}{3} $$ $$ \theta_2 = 26^\circ\text{C} $$