Question

A 6-cm-diameter, 13-cm-high canned drink \((\rho=977\) $\left.\mathrm{kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( initially at \)25^{\circ} \mathrm{C}\( is to be cooled to \)5^{\circ} \mathrm{C}$ by dropping it into iced water at \(0^{\circ} \mathrm{C}\). Total surface area and volume of the drink are \(A_{s}=301.6 \mathrm{~cm}^{2}\) and \(V=367.6 \mathrm{~cm}^{3}\). If the heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\), determine how long it will take for the drink to cool to $5^{\circ} \mathrm{C}$. Assume the can is agitated in water, and thus the temperature of the drink changes uniformly with time. (a) \(1.5 \mathrm{~min}\) (b) \(8.7 \mathrm{~min}\) (c) \(11.1 \mathrm{~min}\) (d) \(26.6 \mathrm{~min}\) (e) \(6.7 \mathrm{~min}\)

Short answer

Answer: It takes approximately 8.76 minutes for the canned drink to cool down to 5°C when placed in iced water.

Step-by-step solution

Convert the given values to SI units

To simplify our calculations, convert all values to the SI (International System of Units) system. Diameter: \(6 cm = 0.06 m\) Total surface area: \(A_{s} = 301.6 cm^2 = 0.03016 m^2\) Volume: \(V = 367.6 cm^3 = 3.676\times 10^{-4} m^3\)

Determine the mass of the drink

Mass, m, can be determined by the following formula: \(m = \rho V\) Here, \(\rho = 977 kg/m^3\) (density) and \(V = 3.676\times 10^{-4} m^3\) (volume) \(m = 977 \times 3.676\times 10^{-4} = 0.359 kg\)

Calculate the time constant

To find out the time constant, τ, we first need to determine the Biot number, \(Bi\), using the following formula: \(Bi = \frac{h L_c}{k}\) where \(L_c = \frac{V}{A_s}\) is the characteristic length, \(h=120 W/m^2K\) is the heat transfer coefficient, and \(k = 0.607 W/mK\) is the thermal conductivity. \(L_c = \frac{3.676\times 10^{-4}}{0.03016} = 0.0122 m\) \(Bi = \frac{120\times 0.0122}{0.607} = 2.414\) Since the Biot number (\(Bi>0.1\)), we cannot directly use the lumped-capacitance method, so we transition to the transient heat conduction approach and calculate the time constant by: \( τ = \frac{ρVc_p}{hA_s}\) where \(ρ = 977 kg/m^3\), \(V= 3.676\times 10^{-4} m^3\), \(c_p = 4180 J/kgK\), \(h = 120 W/m^2K\), and \(A_s = 0.03016 m^2\). \(τ = \frac{977 \times 3.676\times 10^{-4} \times 4180}{120 \times 0.03016} = 452.7 s\)

Calculate the time required for cooling

Using the time constant from the previous step, we can calculate the cooling time using the exponential decay formula: \(T(t) = T_\infty + (T_0 - T_\infty) e^{-t/τ}\) where \(T(t)\) is the temperature at time \(t\), \(T_\infty = 0^{\circ}C\) (temperature of the iced water), \(T_0 = 25^{\circ}C\) (initial temperature of the drink), \(t\) is the time that the drink is in the iced water and \(τ = 452.7 s\) is the time constant. We want to find the time \(t\) when \(T(t) = 5^{\circ}C\). Solving the equation for \(t\), we get: \(5 = 0 + (25 - 0) e^{-t/452.7}\) Dividing both sides by 25 gives: \(\frac{1}{5} = e^{-t/452.7}\) Take the natural logarithm of both sides: \(ln(0.2) = -t / 452.7\) Now, solve for \(t\): \(t = -452.7 ln(0.2) = 525.7 s\) Convert seconds to minutes: \(t = \frac{525.7}{60} = 8.76 min\) So, the answer is close to option (b) \(8.7\mathrm{~min}\).