Answer:[tex]2.89\approx 2.9^{\circ}C/s[/tex]
Explanation:
Given
[tex]Power\left ( P\right )=150 MW[/tex]
mass of core[tex]\left ( m\right )=1.60\times 10^5 kg[/tex]
Average specific heat [tex]\left ( C\right )=0.3349 KJ/kg^{\circ}C[/tex]
And rate of increase of temperature =[tex]\frac{\mathrm{d}T}{\mathrm{d} t}[/tex]
Now
P=[tex]mc\frac{\mathrm{d}T}{\mathrm{d} t}[/tex]
[tex]150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}[/tex]
Thus [tex]\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}[/tex]
[tex]\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s[/tex]