Question

A \(10-\mathrm{cm}\)-inner-diameter, 30-cm-long can filled with water initially at \(25^{\circ} \mathrm{C}\) is put into a household refrigerator at $3^{\circ} \mathrm{C}\(. The heat transfer coefficient on the surface of the can is \)14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming that the temperature of the water remains uniform during the cooling process, the time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is (a) \(0.55 \mathrm{~h}\) (b) \(1.17 \mathrm{~h}\) (c) \(2.09 \mathrm{~h}\) (d) \(3.60 \mathrm{~h}\) (e) \(4.97 \mathrm{~h}\)

Short answer

Answer: (Insert your calculated time here) Check the provided options and select the one that closely matches your answer.

Step-by-step solution

Calculate the heat transfer rate

For the heat transfer process in the can, we can use Newton's law of cooling given by: \(Q = h\cdot A\cdot \Delta T\) Where \(Q\) is the heat transfer rate, \(A\) is the surface area of the can, \(\Delta T\) is the temperature difference, and \(h\) is the heat transfer coefficient. Here, we're given \(h = 14~ \mathrm{W} / \mathrm{m}^{2}\cdot \mathrm{K}\) and the dimensions of the can. To find \(A\), we can calculate the lateral surface area of the can using its diameter and length: \(A = \pi \cdot d \cdot L = \pi \cdot 0.1 \,\mathrm{m} \cdot 0.3 \, \mathrm{m}\) Then we calculate the temperature difference as \(\Delta T = 25 - 3 = 22\,\mathrm{K}\) Now, plug in these values into the heat transfer equation to find \(Q\): \(Q = h\cdot A\cdot \Delta T\)

Use the heat transfer rate to find the time it takes

To calculate the time it takes for the temperature of the water to drop to 5°C, we first need the mass of the water inside the can. We can calculate the volume of the water as: \(V = \pi \cdot \frac{d^2}{4} \cdot L = \pi \cdot \frac{0.1^2}{4} \cdot 0.3 \, \mathrm{m}^3\) Knowing the density of water, we can find the mass: \(m = \rho \cdot V = 1000 \,\mathrm{kg} / \mathrm{m}^{3} \cdot V\) The amount of heat that needs to be removed for the temperature of the water to decrease from 25°C to 5°C can be determined using the specific heat of water, \(c_{p} = 4.18 \,\mathrm{kJ/kg} \cdot \mathrm{K}\): $$Q_{total} = m \cdot c_{p} \cdot \Delta T' = m \cdot c_{p} \cdot (25 - 5)$$ Using Newton's Law of Cooling, the time it takes can be determined as: $$time = \frac{Q_{total}}{Q}$$ Calculate the time using the obtained values and compare it with the options provided.

Check your answer with the given options

Once you have calculated the time, compare it with the options given and select the closest one. If your answer matches one of them, it should be the correct answer.