Consider a cold aluminium canned drink that is initially at a uniform temperature of 3°C. The can is 12.5 cm high and has a diameter of 6 cm. The combined convection/radiation heat transfer coefficient between the can and the surrounding air at 25°C is 10 W/m2·°C. In an effort to slow down the warming of the cold drink, a person puts the can in a perfectly fitting 1-cm-thick cylindrical rubber insulator (k = 0.13 W/m ·°C). How long will it take approximately for the temperature of the drink to rise from 3 °C to 10 °C? (Also see the figure attached)
Key Assumptions:
1 - Assume the top of the can is not covered
2 - For the drink, you can use the properties of water at room temperature, ? = 1000 kg/m3 and Cp = 4180 J/kg.°C.
3 - This is transient heat conduction, and the rate of heat transfer will decrease as the drink warms up and the temperature difference between the drink and the surroundings decreases. However, you can solve this problem approximately by assuming a constant average temperature of (3+10)/2 = 6.5°C for the drink during the process.
A) About 32 min
B) About 43 min
C) About 56 min
D) About 22 min
Please provide an explanation for your answer