Answer:
Explanation:
Given
Initial Temperature of Drink [tex]T_i=25^{\circ}[/tex]
Final Temperature of Drink [tex]T_f=8^{\circ}[/tex]
Temperature of surrounding [tex]T_{\infty }=0^{\circ}[/tex]
Volume of drink [tex]V=335\ ml[/tex]
diameter of container [tex]d=6.5\ cm[/tex]
height of container [tex]h=12.5\ cm[/tex]
Surface area of Container [tex]A=\pi dh+2\times \frac{\pi d^2}{4}[/tex]
[tex]A=321.62\ cm^2[/tex]
Assuming zero temperature gradient
[tex]\frac{T_f-T_{\infty }}{T_i-T_{\infty }}=e^{-\frac{-hA}{\rho Vc}\cdot t}[/tex]
where [tex]h=heat\ transfer\ coefficient\ (170\ W/m^2-^{\circ}C)[/tex]
[tex]c=heat\ capacity\ of\ water(4180\ J/kg-^{\circ}C)[/tex]
substituting values we get
[tex]\frac{8-0}{25-0}=e^{\frac{170\times 321.62\times 10^{-4}}{10^3\times 4180\times 335\times 10^{-6}}\cdot t}[/tex]
[tex]\frac{8}{25}=e^{-3.904\times 10^{-3}t}[/tex]
Taking natural log
[tex]1.1394=3.904\times 10^{-3}t[/tex]
[tex]t=291.85\ s[/tex]
[tex]t=4.86\ min[/tex]
