Question

A 10-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: It takes approximately 1.17 hours for the water temperature to drop to 5°C.

Step-by-step solution

Listing known information

We have been given the following data: - Inner diameter of the can: \(d = 10\) cm - Length of the can: \(L = 30\) cm - Initial water temperature: \(T_{i} = 25^{\circ}\)C - Final water temperature: \(T_{f} = 5^{\circ}\)C - Refrigerator temperature: \(T_{r} = 3^{\circ}\)C - Heat transfer coefficient: \(h = 14\) W/m²K.

Convert units to SI system

We need to convert the given dimensions to meters: - Inner diameter: \(d = \frac{1}{100}\) m - Length: \(L = 0.3\) m

Calculate surface area of the can

We can consider the can as a cylinder and calculate its surface area. The formula for the surface area of a cylinder \(A_{cylinder}\) is given by: $$A_{cylinder} = 2\pi r_{cylinder}(r_{cylinder} + h_{cylinder})$$ where \(r_{cylinder} = \frac{d}{2}= 0.05\) m \(h_{cylinder} = L = 0.3\) m Substitute the values: $$A_{cylinder} = 2\pi(0.05)(0.05 + 0.3) = 0.035 \mathrm{m}^{2}$$

Determine mass and specific heat of the water

To find the mass of the water, we can first find the volume of the cylinder and then use the density of water (ρ) to calculate its mass: $$V_{cylinder} = \pi r_{cylinder}^{2}h_{cylinder} = \pi (0.05)^{2}(0.3)$$ $$m_{water} = \rho V_{cylinder} = 1000 \times \pi (0.05)^{2}(0.3)$$ The specific heat of water \(c_{\mathrm p_{water}}\) is 4.18 kJ/(kg·K) or 4180 J/(kg·K).

Apply Newton's law of cooling

Newton's law of cooling states that the rate of heat loss is proportional to the difference in temperature between the object and its surrounding environment. We can write it as: $$Q = H \Delta T t$$ Where \(Q\) is the heat lost by the water, \(H\) is the heat transfer coefficient multiplied by the surface area of the can, \(\Delta T\) is the difference between the initial and final temperature of the water, and \(t\) is the time required for the cooling process. The heat lost by the water can also be calculated as: $$Q = m c_\mathrm{p} \Delta T_{water}$$ Equating the two expressions for \(Q\) and solving for \(t\): $$H \Delta T t = m c_\mathrm{p} \Delta T_{water}$$ $$t = \frac{m c_\mathrm{p} \Delta T_{water}}{H \Delta T}$$ Substitute the values: $$t = \frac{1000 \times \pi (0.05)^{2}(0.3) \times 4180 (25-5)}{14 \times 0.035 (25-3)} = 1.17 \ \mathrm{h}$$

Conclusion

The time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is approximately 1.17 hours, so the correct answer would be option (b) \(1.17 \mathrm{~h}\).

Key concepts

Newton's Law of Cooling

Newton's Law of Cooling describes how the temperature of an object changes when it is surrounded by a different temperature environment. The rate at which heat is transferred depends mainly on the temperature difference between the object and its surroundings. The larger the temperature difference, the faster the rate of heat transfer.
For this exercise, Newton's Law of Cooling is used to find how long it takes for the water in the can to cool. The equation can be expressed as:
- The rate of heat loss is proportional to the difference in temperature.
- This is mathematically expressed as \( Q = H \Delta T t \), where \(Q\) is the heat lost, \(H\) is the heat transfer coefficient times the surface area, \(\Delta T\) is the temperature difference, and \(t\) is time.
By understanding this law, you can predict cooling times in various scenarios, which is very useful when designing cooling systems or conducting thermal analysis.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. Each material has its own specific heat capacity, which determines how it heats up or cools down.
The water in this exercise has a specific heat capacity of 4180 J/(kg·K). This means that it takes 4180 joules to raise the temperature of 1 kilogram of water by 1 degree Celsius.
- Understanding specific heat capacity helps in calculating how much energy is needed to change the temperature of a substance.
It plays a critical role in heat transfer analysis, such as predicting the cooling period needed in this exercise.

Cylinder Surface Area Calculation

To determine how much heat is exchanged between an object and its surroundings, you often need the object's surface area. For a cylinder, this involves calculating the lateral area and the area of the two circular ends.
The formula for the total surface area \( A_{cylinder} \) of a cylinder is:
\[ A_{cylinder} = 2\pi r_{cylinder} (r_{cylinder} + h_{cylinder}) \]
- Where \( r_{cylinder} \) is the radius and \( h_{cylinder} \) is the height.
In this exercise, the cylinder (can of water) has surface area \(0.035\ m^2\), calculated by using its radius and height. This calculated area is crucial because it directly affects how the heat is transferred to or from the object.

Thermal Conductivity

Thermal conductivity is a measure of how well a material conducts heat. Higher thermal conductivity means that heat can pass through the material quickly.
In the context of this cooling process, the metal can surrounding the water could be made from a material with a high thermal conductivity to facilitate faster heat loss from the water to the refrigerator's cooler environment.
- It is important to choose materials with the right thermal conductivity in applications to ensure efficient temperature control.
While not directly calculated in this particular exercise, thermal conductivity is a background factor in understanding how the can helps in cooling the water faster.

Refrigeration Process

Refrigeration is the process of removing heat from a space or substance to lower its temperature. This exercise acts as a perfect example of how refrigeration works in everyday life.
The refrigerator, being at \(3^\circ C\), absorbs heat from the can of water at \(25^\circ C\). The heat is removed from the water until it reaches its new equilibrium at \(5^\circ C\).
- Refrigerators are designed to maintain a lower internal temperature, which ensures effective food preservation.
Understanding this process is essential for those learning about heat transfer, as it exemplifies how thermal systems are designed to manage temperature efficiently.