(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 J of work per cycle per cylinder. If the engine runs at 25 cycles per second (1500 rpm), determine (a) the work done per second, and (b) the total heat input per second from the gasoline. (c) If the energy content of gasoline is 120 MJ per gallon, how far can the car go on one gallon (or on one liter), assuming 1500 rpm in high gear corresponds to a speed of 80 km/h?
(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 J of work per cycle per cylinder. If the engine runs at 25 cycles per second (1500 rpm), determine (a) the work done per second, and (b) the total heat input per second from the gasoline. (c) If the energy content of gasoline is 120 MJ per gallon, how far can the car go on one gallon (or on one liter), assuming 1500 rpm in high gear corresponds to a speed of 80 km/h?
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![**Problem 6**
(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 J of work per cycle per cylinder. If the engine runs at 25 cycles per second (1500 rpm), determine:
(a) The work done per second, and
(b) The total heat input per second from the gasoline.
(c) If the energy content of gasoline is 120 MJ per gallon, how far can the car go on one gallon (or on one liter), assuming 1500 rpm in high gear corresponds to a speed of 80 km/h?
**Detailed explanations:**
- **(a) The work done per second:**
Since the engine operates with four cylinders and each cylinder delivers 180 J of work per cycle, and the engine runs at 25 cycles per second:
Work done per second \(W\) is calculated by:
\[
W = \text{work per cycle per cylinder} \times \text{number of cylinders} \times \text{cycles per second}
\]
\[
W = 180 \, \text{J} \times 4 \times 25 = 18000 \, \text{J/s}
\]
- **(b) The total heat input per second:**
Since the efficiency (η) of the engine is 0.22:
\[
\eta = \frac{\text{useful work output}}{\text{total heat input}}
\]
Rearranging to find the total heat input \(Q\):
\[
Q = \frac{\text{useful work output}}{\eta} = \frac{18000 \, \text{J/s}}{0.22} \approx 81818 \, \text{J/s}
\]
- **(c) Distance traveled on one gallon of gasoline:**
Given:
- Energy content of gasoline = 120 MJ per gallon.
- \(1 \, \text{MJ} = 10^6 \, \text{J}\).
Total energy content in one gallon:
\[
E_{\text{total}} = 120 \times 10^6 \, \text{J}
\]
Total work done by the engine using this energy:
\[
\text{Total useful work} = \eta \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Facfb296d-c42b-4e28-876a-1baf5375f9c9%2F6d404db3-bd25-4f0d-a6e8-9befe89c49be%2Fxkpqgi_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 6**
(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 J of work per cycle per cylinder. If the engine runs at 25 cycles per second (1500 rpm), determine:
(a) The work done per second, and
(b) The total heat input per second from the gasoline.
(c) If the energy content of gasoline is 120 MJ per gallon, how far can the car go on one gallon (or on one liter), assuming 1500 rpm in high gear corresponds to a speed of 80 km/h?
**Detailed explanations:**
- **(a) The work done per second:**
Since the engine operates with four cylinders and each cylinder delivers 180 J of work per cycle, and the engine runs at 25 cycles per second:
Work done per second \(W\) is calculated by:
\[
W = \text{work per cycle per cylinder} \times \text{number of cylinders} \times \text{cycles per second}
\]
\[
W = 180 \, \text{J} \times 4 \times 25 = 18000 \, \text{J/s}
\]
- **(b) The total heat input per second:**
Since the efficiency (η) of the engine is 0.22:
\[
\eta = \frac{\text{useful work output}}{\text{total heat input}}
\]
Rearranging to find the total heat input \(Q\):
\[
Q = \frac{\text{useful work output}}{\eta} = \frac{18000 \, \text{J/s}}{0.22} \approx 81818 \, \text{J/s}
\]
- **(c) Distance traveled on one gallon of gasoline:**
Given:
- Energy content of gasoline = 120 MJ per gallon.
- \(1 \, \text{MJ} = 10^6 \, \text{J}\).
Total energy content in one gallon:
\[
E_{\text{total}} = 120 \times 10^6 \, \text{J}
\]
Total work done by the engine using this energy:
\[
\text{Total useful work} = \eta \
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