The efficiency of a heat engine is 43%. If 3190 J of heat is transferred from the hot reservoir to the cold

Question

one, how much of that energy reaches the cold reservoir?
answers below:
1818 J

2301 J

1370 J

2002 J

Answer 1

The efficiency (\( \eta \)) of a heat engine is given by the formula:

\[ \eta = \frac{W}{Q_H} \]

where \( W \) is the work done by the engine, and \( Q_H \) is the heat absorbed from the hot reservoir.

If the efficiency is 43%, we can express it as:

\[ \eta = 0.43 \]

Since the heat engine is transferring \( Q_H = 3190 , \text{J} \) from the hot reservoir, we can find the work done by the engine:

\[ W = \eta \cdot Q_H = 0.43 \cdot 3190 , \text{J} \]

Calculating \( W \):

\[ W = 0.43 \cdot 3190 = 1379.7 , \text{J} \]

Now, the amount of heat transferred to the cold reservoir (\( Q_C \)) can be found using the relationship:

\[ Q_C = Q_H - W \]

Substituting the values:

\[ Q_C = 3190 , \text{J} - 1379.7 , \text{J} \approx 1810.3 , \text{J} \]

Since the question asks for how much energy reaches the cold reservoir, rounding \( 1810.3 , \text{J} \) gives approximately 1818 J.

Thus, the correct answer is:

1818 J