Step 1: Understanding the Concept:
Efficiency (\( \eta \)) of a heat engine is defined as the ratio of the mechanical work output (\( W \)) to the heat energy input (\( Q \)). : Key Formula or Approach:
1. \( \eta = \frac{W}{Q} \implies W = \eta \times Q \).
2. Conversion: \( 1 \text{ cal} \approx 4.2 \text{ J} \). Step 2: Detailed Explanation:
Given data:
Efficiency \( \eta = 1/3 \).
Heat input \( Q = 1 \text{ kcal} = 1000 \text{ cal} \).
First, calculate the work in calories:
\[ W = \eta \times Q = \frac{1}{3} \times 1000 \text{ cal} = 333.3 \text{ cal} \]
Since the options are in Joules and Calories, let's convert the heat input to Joules:
\[ Q = 1000 \text{ cal} \times 4.2 \text{ J/cal} = 4200 \text{ J} \]
Now, calculate the work output in Joules:
\[ W = \eta \times Q = \frac{1}{3} \times 4200 \text{ J} = 1400 \text{ J} \] Step 3: Final Answer:
The work the engine can perform is 1400 J.
