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If it takes about \(335 \mathrm{~kJ}\) to freeze \(1 \mathrm{~kg}\) of water, how much ice could an ice maker having a 1 -ton refrigeration capacity produce in 24 hours?

Short Answer

Expert verified
About 10,496.57 kg of ice can be produced in 24 hours.

Step by step solution

01

Understand the Units

Identify the units involved in the problem. The energy required to freeze water is given in kilojoules \(\text{kJ}\) and the capacity of the ice maker is given in tons. First, we need to understand that 1 ton of refrigeration capacity is equivalent to \( 3.51685 \times 10^6 \) \(\text{kJ/day}\).
02

Calculate the Total Energy Output

Calculate the total energy output of the ice maker in one day. Since one ton of refrigeration is \( 3.51685 \times 10^6 \) \(\text{kJ/day}\), \(\text{Total Energy Output} = 3.51685 \times 10^6 \) \(\text{kJ}\).
03

Determine the Energy Required per kg of Ice

According to the problem, it takes about \( 335 \) \(\text{kJ}\) to freeze \( 1 \) \(\text{kg}\) of water.
04

Calculate the Amount of Ice Produced

Now, use the total energy output and the energy required per kg of ice to find out how much ice can be produced. \(\text{Amount of Ice} = \frac{ \text{Total Energy Output} }{ \text{Energy per kg} } = \frac{3.51685 \times 10^6 }{ 335 } \)
05

Perform the Division

Perform the division to find the amount of ice produced. \(\frac{3.51685 \times 10^6 }{ 335 } \approx 10,496.57 \) kg.

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Key Concepts

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

Energy conversion
Understanding energy conversion is crucial in thermodynamics. Energy conversion refers to changing energy from one form to another. In the given exercise, we convert the energy capacity of refrigeration (measured in tons) into kilojoules (kJ). This conversion helps us understand the total energy an ice maker can output in a day. One ton of refrigeration equals 3,516,850 kJ per day. Energy can be converted into other forms, such as mechanical work or heat. This understanding is vital for applications like refrigeration and heating, where energy transformation is necessary to achieve the desired outcomes.
Refrigeration capacity
Refrigeration capacity is the rate at which a refrigeration system can remove heat from a space. For ice makers, we measure this capacity in tons. One ton of refrigeration refers to the energy required to freeze 1 ton (2,000 pounds) of water at 32°F (0°C) in one day. This measure equals 3,516,850 kJ per day. Using this measurement, we can calculate how much ice an ice maker can produce within a specific time frame. For instance, in the exercise, knowing the ice maker's 1-ton capacity helps us compute the total energy output and subsequently calculate the amount of ice it can produce over 24 hours.
Phase change
A phase change involves the transition of a substance from one state of matter to another. In the context of the exercise, we focus on the phase change from liquid water to solid ice. The energy requirement for this transition is known as the latent heat of fusion. For water, this is 335 kJ/kg, meaning it takes 335 kilojoules to freeze 1 kilogram of water. Phase changes are essential concepts in thermodynamics, as they explain energy requirements and temperature changes during transformations. Understanding phase changes helps in calculating energy usage in processes like freezing, boiling, or condensation.

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Most popular questions from this chapter

You see an advertisement for a natural gas-fired absorption refrigeration system. How can burning natural gas play a role in achieving cooling?

A vapor-compression refrigeration system with a capacity of 10 tons has superheated Refrigerant 134 a vapor entering the compressor at \(15^{\circ} \mathrm{C}, 4\) bar, and exiting at 12 bar. The compression process can be modeled by \(p v^{1.01}=\) constant. At the condenser exit, the pressure is \(11.6\) bar, and the temperature is \(44^{\circ} \mathrm{C}\). The condenser is water-cooled, with water entering at \(20^{\circ} \mathrm{C}\) and leaving at \(30^{\circ} \mathrm{C}\) with a negligible change in pressure. Heat transfer from the outside of the condenser can be neglected. Determine (a) the mass flow rate of the refrigerant, in \(\mathrm{kg} / \mathrm{s}\). (b) the power input and the heat transfer rate for the compressor, each in \(\mathrm{kW}\). (c) the coefficient of performance. (d) the mass flow rate of the cooling water, in \(\mathrm{kg} / \mathrm{s}\). (e) the rates of exergy destruction in the condenser and expansion valve, each expressed as a percentage of the power input. Let \(T_{0}=20^{\circ} \mathrm{C}\).

\(10.34\) Air at 2 bar, \(380 \mathrm{~K}\) is extracted from a main jet engine compressor for cabin cooling. The extracted air enters a heat exchanger where it is cooled at constant pressure to \(320 \mathrm{~K}\) through heat transfer with the ambient. It then expands adiabatically to \(0.95\) bar through a turbine and is discharged into the cabin. The turbine has an isentropic efficiency of \(75 \% .\) If the mass flow rate of the air is \(1.0 \mathrm{~kg} / \mathrm{s}\), determine (a) the power developed by the turbine, in \(\mathrm{kW}\). (b) the rate of heat transfer from the air to the ambient, in \(\mathrm{kW}\).

An ideal vapor-compression heat pump cycle with Refrigerant \(134 \mathrm{a}\) as the working fluid provides heating at a rate of \(15 \mathrm{~kW}\) to maintain a building at \(20^{\circ} \mathrm{C}\) when the outside temperature is \(5^{\circ} \mathrm{C}\). Saturated vapor at \(2.4\) bar leaves the evaporator, and saturated liquid at 8 bar leaves the condenser. Calculate (a) the power input to the compressor, in \(\mathrm{kW}\). (b) the coefficient of performance. (c) the coefficient of performance of a Carnot heat pump cycle operating between thermal reservoirs at 20 and \(5^{\circ} \mathrm{C}\).

Explain how a household refrigerator can be viewed as a heat pump that heats the kitchen. If you knew the refrigerator's coefficient of performance, could you calculate its coefficient of performance when viewed as a heat pump?

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