Question

During a picnic on a hot summer day all thecold drinks disappeared quickly, and the onlyavailable drinks were those at ambienttemperature of 25ºC. In an effort to cool a 335-ml drink in a can, which is 12.5 cm high andhas a diameter of 6.5cm, a person grabs thecan and starts shaking it in the iced water of thechest at 0ºC. The temperature of the drink canbe assumed to be uniform at all time, and theheat transfer coefficient between the iced waterand the aluminum can is 170 W/m2-ºC. Usingthe properties of water for the drink, estimatehow long it will take for the canned drink to coolto 8ºC.

Answer 1

Step-by-step explanation:

Given

Initial Temperature of Drink
`T_i=25^(\circ)`

Final Temperature of Drink
`T_f=8^(\circ)`

Temperature of surrounding
`T_(\infty )=0^(\circ)`

Volume of drink
`V=335\ ml`

diameter of container
`d=6.5\ cm`

height of container
`h=12.5\ cm`

Surface area of Container
`A=\pi dh+2* (\pi d^2)/(4)`

`A=321.62\ cm^2`

Assuming zero temperature gradient

`(T_f-T_(\infty ))/(T_i-T_(\infty ))=e^{-(-hA)/(\rho Vc)\cdot t}`

where
`h=heat\ transfer\ coefficient\ (170\ W/m^2-^(\circ)C)`

`c=heat\ capacity\ of\ water(4180\ J/kg-^(\circ)C)`

substituting values we get

`(8-0)/(25-0)=e^{(170* 321.62* 10^(-4))/(10^3* 4180* 335* 10^(-6))\cdot t}`

`(8)/(25)=e^{-3.904* 10^(-3)t}`

Taking natural log

`1.1394=3.904* 10^(-3)t`

`t=291.85\ s`

`t=4.86\ min`