Answer:
a) [tex]Q_{in} = 13.742\,kW[/tex], b) [tex]\Delta S = 370.15\,\frac{kJ}{K}[/tex]
Explanation:
a) The heat transfered to the egg is computed by the First Law of Thermodynamics:
[tex]Q_{in} +U_{sys,1} - U_{sys,2} = 0[/tex]
[tex]Q_{in} = U_{sys,2} - U_{sys,1}[/tex]
[tex]Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})[/tex]
[tex]Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)[/tex]
[tex]Q_{in} = 13.742\,kW[/tex]
b) The amount of entropy generation is determined by the Second Law of Thermodynamics:
[tex]\Delta S = \frac{Q_{in}}{T_{in}}[/tex]
[tex]\Delta S = \frac{13.742\,kJ}{370.15\,K}[/tex]
[tex]\Delta S = 370.15\,\frac{kJ}{K}[/tex]
