Chapter 15, Problem 53GP

(a)

To determine

The work done per second.

Answer to Problem 53GP

The work done per second is $4\times {10}^4\text{J}/\text{s}$ .

Explanation of Solution

Given:

The given efficiency is $0.25$ and delivers $220\text{J}$ of work per cycle per cylinder. The engine fires at $45$ cycles per second.

Formula used:

Applying work done formula,

  $W=\frac{220J}{\left(cycle\right)\left(cylinder\right)}$

  $W$ is a work done of engine.

Calculation:

Calculate the work done,

  $\begin{array}{c}W=\frac{220\text{J}}{\left(\text{cycle}\right)\left(\text{cylinder}\right)}\\ \frac{W}{s}=\left(\frac{220\text{J}}{\left(\text{cycle}\right)\left(\text{cylinder}\right)}\right)\left(\frac{45\text{cycle}}{\text{s}}\right)\left(4\text{cylinders}\right)\\ \frac{W}{s}=3.96\times {10}^4\text{J/s}\\ \approx 4\times {10}^4\text{J/s}\end{array}$

Conclusion:

Therefore, the work done per second is $4\times {10}^4\text{J}/\text{s}$ .

(b)

To determine

The total heat input per second from the fuel.

Answer to Problem 53GP

The total heat input per second from the fuel is $1.6\times {10}^5\text{J}/\text{s}$ .

Explanation of Solution

Given:

The given efficiency is $0.25$ and delivers $220\text{J}$ of work per cycle per cylinder. The engine fires at $45$ cycles per second.

Formula used:

The efficiency of heat formula,

  $e=\frac{W}{Q_H}$

  $e$is a efficiency of engine, $W$ is work done of engine and $Q_H$ is a heat of engine.

Calculation:

Substitute the values in the formula

  $\begin{array}{c}e=\frac{W}{Q_H}\\ Q_H=\frac{W}{e}\\ =\frac{4\times {10}^4\text{J}/\text{s}}{0.25}\\ =1.6\times {10}^5\text{J}/\text{s}\end{array}$

Therefore, the total heat input per second from the fuel is $1.6\times {10}^5\text{J}/\text{s}$ .

(c)

To determine

The energy of one liter.

Answer to Problem 53GP

The energy content of gasoline in $35\text{MJ}$ per liter is $3.7\text{min}$ .

Explanation of Solution

Given:

The energy content of gasoline in $35\text{MJ}$ per liter.

Calculation:

Calculate the time of one liter gasoline will work,

  $\begin{array}{c}\left(\frac{3.5\times {10}^7\text{J}}{\text{L}}\right)\left(1\text{L}\right)\left(\frac{8}{1.6\times {10}^5\text{J}}\right)=2.1875\times {10}^2\text{s}\\ 219\text{s}=3.65\text{min}\\ \approx 3.7\min \end{array}$

Conclusion:

Therefore, the energy content of gasoline in $35\text{MJ}$ per liter is $3.7\min$ .