Aflam masa apei:
[tex]Q=mc\Delta T\ \ \ \ \Rightarrow \ \ \ \ m=\dfrac{Q}{c\Delta T}[/tex]
Aflam volumul de apa:
[tex]\rho=\dfrac{m}{V}\ \ \ \ \Rightarrow \ \ \ \ [/tex][tex]V=\dfrac{m}{\rho}=\dfrac{Q}{\rho c\Delta T}[/tex]
Dar volumul este aria bazei ori inaltimea: [tex]V=Sh[/tex]
Rezulta:
[tex]S=\dfrac{V}{h}=\dfrac{Q}{\rho h c\Delta T}.[/tex]
Aria unui cerc este: [tex]S=\pi R^2[/tex]
Rezulta:
[tex]R=\sqrt{\dfrac{S}{\pi}}=\sqrt{\dfrac{Q}{\pi\rho h c\Delta T}}[/tex].
Datele problemei:
[tex]Q=3937.59\ J \\ \Delta T=100\\ h=0.30 \ m\\ \rho=\text{cauta pe net densitatea apei}\\ c=\text{cauta caldura specifica a apei}[/tex]
