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Your next-door neighbor, Josephine Rackstraw, surprised her husband last January by having a hot tub installed in their back yard while he was away on an ice-fishing trip. It surprised him, all right, but instead of being pleased he was horrified. "Have you lost your mind, Josephine?" he sputtered. "It will cost a fortune to keep this thing hot, and you know what the President said about conserving energy." "Don't be silly, Ralph," she replied. "It can't cost more than a few pennies a day, even in the dead of winter." "No way -just because you have a PhD, you think you're an expert on everything!" They argued for a while, bringing up several issues that each had been storing for just such an occasion. After calming down and using the tub for a week, they remembered their neighbor (i.e., you) had a chemical engineering education and came to ask if you could settle their argument. You asked a few questions, made several observations, converted everything to metric units, and arrived at the following data, all corresponding to an average outside temperature of \(5^{\circ} \mathrm{C}\). \- The tub holds 1230 liters of water. \- Ralph normally keeps the tub temperature at \(29^{\circ} \mathrm{C}\), raises it to \(40^{\circ} \mathrm{C}\) when he plans to use it, keeps it at \(40^{\circ} \mathrm{C}\) for about one hour, and drops it back to \(29^{\circ} \mathrm{C}\) when he is finished. \- During heating, it takes about three hours for the water temperature to rise from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). When the heat is shut off, it takes eight hours for the water temperature to drop back to \(29^{\circ} \mathrm{C}\). \- Electricity costs 10 cents per kilowatt-hour. Taking the heat capacity of the tub contents to be that of pure liquid water and neglecting evaporation, answer the following questions. (a) What is the average rate of heat loss ( \(k W\) ) from the tub to the outside air? (Hint: Consider the period when the tub temperature is dropping from \(40^{\circ} \mathrm{C}\) to \(29^{\circ} \mathrm{C}\).) (b) At what average rate ( \(\mathrm{kW}\) ) does the tub heater deliver energy to the water when raising the water temperature? What is the total quantity of electricity (kW\cdoth) that the heater must deliver during this period? [Consider the result of Part (a) when performing the calculation.] (c) (These answers should settle the argument.) Consider a day in which the tub is used once. Use the results of Parts (a) and (b) to estimate the cost (S) of heating the tub from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) and the cost \((\mathrm{S})\) of keeping the tub at a constant temperature. (There is no cost for the period in which \(T\) is dropping.) What is the total daily cost of running the tub? Assume the rate of heat loss is independent of the tub temperature. (d) The tub lid, which is an insulator, is removed when the tub is in use. Explain how this fact would probably affect your cost estimates in Part (c).

Step by step solution

01

Calculate heat transferred

The heat transferred from the hot tub to the surrounding can be calculated by applying the specific heat formula: Q = mcΔT, where m = mass, c = specific heat capacity of water (4.186 kJ/kg°C), ΔT = change in temperature. Convert the volume of water (1230 litres) to mass using the fact that 1 litre of water weighs approximately 1 kg, so m = 1230 kg. The change in temperature (ΔT) is 40°C - 29°C = 11°C.

02

Calculate average rate of heat loss

The average rate of heat loss (P = Q/t) can be calculated by dividing the heat transferred (Q) by the time taken for the water temperature to drop (t), which is 8 hours. Convert this time to seconds for consistency in units.

03

Calculate the average rate of energy delivery

To calculate the average rate of energy delivery to the water, you need to account for both the energy required to heat the water (calculated in Step 1) and the energy lost to the outside atmosphere during the heating process (calculated in Step 2). It takes 3 hours to heat the water; by converting this time to seconds and adding to the loss rate, you'll get the average rate of energy delivery (Pt = Ph + P).

04

Calculate the total amount of electricity

The total amount of electricity can be calculated by multiplying the rate of energy delivery by the time it took to heat the water up (which is 3 hours, converted to seconds). Then, convert this energy from Joules to kWh by dividing by 3.6 million (as 1 kWh is 3.6 million Joules).

05

Calculate the cost

Using the cost of electricity (10 cents per kWh), calculate the total cost of the heat energy for maintaining the temperature and for raising the temperature. Once these costs are calculated, add them together to get the total daily cost of running the tub.

06

Consider the effects of the removed lid

Without the insulating lid, more heat will be lost to the environment when the tub is in use, increasing the rate of energy consumption. As a result, the cost estimated in step 5 would most likely be an underestimation. However, since this is a qualitative question and doesn't require a calculation, just understand that an increased rate of heat loss means more energy (and thus more cost) will be required to maintain the same temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer Calculations

Understanding heat transfer calculations is essential for solving problems related to maintaining temperature in systems like a hot tub. Heat transfer refers to the movement of thermal energy from one object or substance to another due to a temperature difference. In Josephine's case, quantifying this energy exchange is crucial to resolve the dispute about the cost of keeping the hot tub warm.

To calculate the heat transfer, we use the formula \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass of the water in the tub, \( c \) is the specific heat capacity of water, and \( \Delta T \) is the change in temperature. In our example, the specific heat capacity of water (\( c \) = 4.186 kJ/kg°C) is the amount of energy required to raise the temperature of 1 kilogram of water by 1°C. The mass of the water is equivalent to its volume, given that 1 liter of water approximately weighs 1 kg. Thus, with the volume of 1230 liters, we have \( m = 1230 \) kg.

Breaking down the heat loss involves finding the average rate, which is done by dividing the total heat transfer by the time it takes for the tub to cool down. It's important to convert this time to the appropriate units (seconds) to match the unit of energy commonly used in such calculations (Joules).

Specific Heat Capacity

Specific heat capacity plays a pivotal role in thermal analysis and energy calculations. It is defined as the quantity of heat required to change the temperature of a unit mass of a substance by one degree Celsius. In chemical engineering, this property allows engineers to determine how much energy is required to heat up or cool down a specific amount of material.

In Josephine's hot tub scenario, the specific heat capacity of water (\( c = 4.186 \) kJ/kg°C) ensures that a fairly large amount of energy is needed to alter the temperature. This is due to water’s high specific heat capacity compared to many other substances. As a result, the specific heat capacity is essential in calculating both the energy required to heat the tub and the rate at which this energy is lost to the environment.

Improving textbook solutions on this topic might involve demonstrating how the specific heat capacity affects the time and energy required to raise or lower the temperature of the tub. Providing real-world examples and emphasizing the practical application of the concept can significantly enhance students' comprehension.

Energy Conservation in Processes

Energy conservation is a principle stating that the total energy of an isolated system remains constant. In chemical engineering, and specifically in processes like heating a hot tub, this principle ensures that all energy transfers are accounted for. It's not just about the energy required to heat the tub but also about the heat loss to the surroundings which must be minimized for efficient energy use.

In the context of the hot tub's heat maintenance, energy conservation implicates both the energy supplied by the heater and the subsequent loss when the tub is not in use. The heat loss to the environment, whether through the sides of the tub or when the insulating lid is removed, necessitates additional energy to maintain the desired temperature. Factoring the cost of this energy is critical when evaluating the expense of operating the hot tub, as per Ralph's concerns.

Enhancements in educational solutions could focus on illustrating the concepts of energy efficiency and cost–benefit analysis in household processes. Discussing the impact of insulating materials, comparing different energy sources, and providing strategies for energy conservation can solidify not only the theoretical understanding but also the practical implications of energy conservation in everyday life.