Respuesta :
• Given:
Work done = 1.85 x 10⁸ J
Efficiency = 4%
Let's solve for the following:
• (a). How much heat transfer (in J) occurs to the environment?
Apply the formula:
[tex]\eta=\frac{W}{Q_h}[/tex]Where Qh is the amount of heat transfer.
rewrite the formula for Qh:
[tex]Q_h=\frac{W}{\eta}[/tex]The equation for the work done by the engine will be:
[tex]W=Q_h-Q_c[/tex]Now, substitute W/n for Wh in the equation for the work done:
[tex]\begin{gathered} W=\frac{W}{\eta}-Q_c \\ \text{ } \\ Rewrite\text{ the equation for }Q_c: \\ Q_c=\frac{W}{\eta}-W \\ \\ Q_c=W(\frac{1}{\eta}-1) \end{gathered}[/tex]Where:
n is the efficiency = 4% = 0.04
W is the work done.
Plug in the values and solve for Qc:
[tex]\begin{gathered} Q_c=(1.85\times10^8)(\frac{1}{0.04}-1) \\ \\ Q_c=(1.85\times10^8)(25-1) \\ \\ Q_c=(1.85\times10^8)(24) \\ \\ Q_c=4.44\times10^9\text{ J} \end{gathered}[/tex]Therefore, the amount of heat transfer that occurs in the environment is 4.44 x 10⁹ J.
• Part B:
Given:
Amount of heat produced by each barrel when burned = 6.00 x 10⁹ J.
To find the amount of barrels of fuel consumed, apply the formula:
[tex]N=\frac{Q_c+W}{6.00\times10^9}[/tex]Substitute values in the formula and solve for N, where N is the number of barrels of fuel burned.
[tex]\begin{gathered} N=\frac{(4.44\times10^9)+(1.85\times10^8)}{6.00\times10^9} \\ \\ N=0.771 \end{gathered}[/tex]Therefore, the amount of fuel consumed is 0.771 barrels.
ANSWER:
• (a). 4.44 x 10⁹ J
• (b). 0.771 barrels.
