Question

The engine of a large ship does 2.00×10⁸J of work with an efficiency of 5.00% (a) How much heat transfer occurs to the environment (b) How many barrels of fuels are consumed if each barrel produces 6.00×10⁹J of heat transfer when burned?

Short answer

Amount of heat transferred to the environment is 3.8×10⁹J and .667 barrels of fuels are consumed.

Step-by-step solution

Step: 1 Introduction-

Heat engine is a device by which a system is made to undergo a cyclic process. It results conversion of heat into work. First and second law helps in the operation of Heat Engine.

Step: 2 Given Parameters and formula to be used-

Amount of work done by the engine $W=2.00\times {10}^8\mathrm{J}$

Efficiency $\eta =5.00\%$

Step: 3 Calculations

(a)

Relation to be used

$$\begin{gathered} h=\frac{W}{Q_1} \\ W=Q_1-Q_2 \end{gathered}$$

Here

$$\begin{gathered} h-\mathrm{Efficinency}\mathrm{of}\mathrm{the}\mathrm{heat}\mathrm{Engine} \\ W-\mathrm{Work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{Heat}\mathrm{Engine} \\ {Q}_1-\mathrm{Amount}\mathrm{of}\mathrm{Heat}\mathrm{Extracted}\mathrm{from}\mathrm{Heat}\mathrm{Engine} \\ {Q}_2-\mathrm{Amount}\mathrm{of}\mathrm{Heat}\mathrm{transfered}\mathrm{to}\mathrm{Environment} \end{gathered}$$

$$\begin{gathered} h=\frac{W}{Q_1} \\ {Q}_1=\frac{W}{h} \\ {Q}_1=\frac{2.00\times {10}^8}{.05} \\ {Q}_1=4\times {10}^9\mathrm{J} \end{gathered}$$

Further,

$$\begin{gathered} W=Q_1-Q_2 \\ {Q}_2=Q_1-W \\ {Q}_2=4\times {10}^{9}J-2.00\times {10}^8\mathrm{J} \\ {Q}_2=3.8\times {10}^9\mathrm{J} \end{gathered}$$

(b) Barrels of fuel consumed -

$$\begin{gathered} n=\frac{4\times {10}^9\mathrm{J}}{6\times {10}^9\mathrm{J}} \\ =0.667 \end{gathered}$$

So, amount of heat transferred to the environment is $3.8\times {10}^9\mathrm{J}$and barrels of fuel consumed are $0.667$.