Question

Ce lucru mecanic produce un motor cu randamentul 30% daca consuma 7 litri de benzina q=46 Mj/kg si densitatea 800kg/m cub.
Ce caldura se pierde la mobilul respectiv?

Answer 1

[tex]\tt Se~da:\\ \\ \eta=30\%=0,3\\ \\ V=7l=0,007m^3\\ \\ q=46 \frac{MJ}{kg}=46\times 10^{6}\frac J{kg}\\ \\ \rho=800\frac {kg}{m^3}\\ \\ L=?J\\ \\ Q_{pe}=?J\\ \\ \\[/tex]



[tex]\tt Formule:\\ \\ \eta=\frac L Q\\ \\ \eta=\frac L {m \times q}\\ \\ L=\eta \times V \times \rho \times q\\ \\ \\ Q_{pe}=Q-L\\ \\ Q_{pe}=q\times m-L\\ \\ Q_{pe}=q\times V \times \rho-L \\ \\ \\ Calcule:\\ \\ L=0,3 \times 0,007 \times 800 \times 46 \times 10^6=77,28\times 10^6 J \\ \\ Q_{pe}=46\times 10^6 J \times 0,007 \times 800-77,28\times 10^6 J=180320000J [/tex]