Question

During a picnic on a hot summer day, the only available drinks were those at the ambient temperature of \(90^{\circ} \mathrm{F}\). In an effort to cool a 12 -fluid-oz drink in a can, which is 5 in high and has a diameter of $2.5 \mathrm{in}$, a person grabs the can and starts shaking it in the iced water of the chest at \(32^{\circ} \mathrm{F}\). The temperature of the drink can be assumed to be uniform at all times, and the heat transfer coefficient between the iced water and the aluminum can is $30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. Using the properties of water for the drink, estimate how long it will take for the canned drink to cool to \(40^{\circ} \mathrm{F}\). Solve this problem using lumped system analysis. Is the lumped system analysis applicable to this problem? Why?

Short answer

Question: Estimate the time it will take for a canned drink to cool from 90°F to 40°F in iced water with an ambient temperature of 32°F. Answer: The estimated time using the lumped system analysis is about 21.8 minutes. However, this analysis may not be accurate due to the Biot number being greater than 0.1, and the actual cooling time may be different.

Step-by-step solution

Calculate the surface area of the can

The can is a cylinder with a diameter of 2.5 inches and a height of 5 inches. The surface area of a cylinder can be calculated using the following formula: \(A = 2 \pi \: r(h+r)\) where \(A\) is the surface area, \(r\) is the radius, and \(h\) is the height. We can calculate the surface area as follows: \(r = \dfrac{2.5}{2} = 1.25 \mathrm{in}\) \(A = 2 \pi (1.25)(5+1.25) = 47.12 \: \mathrm{in}^2\) So, the surface area of the can is 47.12 square inches.

Calculate the Biot number

The Biot number (Bi) is a dimensionless quantity that helps us decide if the lumped system analysis is applicable. The Biot number is given by: \(Bi = \dfrac{h_c \: L_c}{k}\) where \(h_c\) is the heat transfer coefficient, \(L_c\) is the characteristic length, and \(k\) is the thermal conductivity of the material. For this problem, \(h_c = 30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^2 \cdot ^{\circ} \mathrm{F}\) and we will use \(L_c = r\) and the thermal conductivity of water \(k = 0.345 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot^{\circ} \mathrm{F}\). First, we need to convert the radius to feet: \(L_c = 1.25 \mathrm{in} \times \dfrac{1 \mathrm{ft}}{12 \mathrm{in}} = 0.1042 \mathrm{ft}\) Now, we can calculate the Biot number: \(Bi = \dfrac{30 \cdot 0.1042}{0.345} = 9.04\) Since \(Bi > 0.1\), the lumped system analysis may not be applicable here. However, let's proceed to step 3 to estimate the cooling time.

Estimate the cooling time using lumped system analysis

The lumped system analysis formula is given by: \(t = \dfrac{\rho \: V \: c_p(T_i - T_\infty)}{h_c \: A(T_i - T_\infty)}\) where \(t\) is the time, \(\rho\) is the density of water (62.43 lb/ft³), \(V\) is the volume of the drink, \(c_p\) is the specific heat capacity of water (1 Btu/lb°F), \(T_i\) is the initial temperature (90°F), \(T_\infty\) is the ambient temperature (40°F), \(h_c\) is the heat transfer coefficient, and \(A\) is the surface area. First, we need to convert the volume of the drink to ft³: \(V = 12 \: \mathrm{fluid} \: \mathrm{oz} \times \dfrac{1 \mathrm{ft}^{3}}{958.5 \: \mathrm{fluid} \: \mathrm{oz}} = 0.01252 \: \mathrm{ft}^{3}\) Now, we can find the cooling time: \(t = \dfrac{62.43 \times 0.01252 \times 1 \times (90 - 32)}{30 \times (47.12 \times \dfrac{1 \mathrm{ft}^{2}}{144 \: \mathrm{in}^{2}}) \times (90 - 32)} = 0.3636 \: \mathrm{h}\) So, it will take approximately 0.3636 hours or about 21.8 minutes for the drink to cool to 40°F using the lumped system analysis. However, keep in mind that the Biot number suggests that this analysis may not be accurate, and the actual cooling time may be different.