Question

For each \(\mathrm{kW}\) of power input to an ice maker at steady state, determine the maximum rate that ice can be produced, in \(\mathrm{kg} / \mathrm{h}\), from liquid water at \(0^{\circ} \mathrm{C}\). Assume that \(333 \mathrm{~kJ} / \mathrm{kg}\) of energy must be removed by heat transfer to freeze water at \(0^{\circ} \mathrm{C}\), and that the surroundings are at \(20^{\circ} \mathrm{C}\).

Short answer

10.81 kg/h

Step-by-step solution

- Identify given values

The power input to the ice maker is 1 kW. The latent heat of fusion for water is given as 333 kJ/kg. The surroundings are at 20°C and the freezing point of water is 0°C.

- Understand the problem

We need to find the maximum rate of ice production in kg/h from liquid water at 0°C with a given power input of 1 kW, while assuming that 333 kJ/kg of energy must be removed to freeze the water.

- Calculate the energy input per hour

First, convert the power input from kW to kJ/h. Since 1 kW = 1 kJ/s, multiply by the number of seconds in an hour (3600 s). 1 kW = 1 kJ/s Energy input per hour = 1 kW * 3600 s = 3600 kJ/h.

- Calculate the mass of ice produced

Using the given latent heat of fusion for water, calculate how much ice can be produced by the energy input. Energy required to freeze 1 kg of water = 333 kJ Maximum mass of ice produced per hour = Energy input per hour / Latent heat of fusion Therefore, mass of ice = 3600 kJ/h / 333 kJ/kg = 10.81 kg/h.

Key concepts

Latent Heat of Fusion

The latent heat of fusion is the amount of energy required to change a substance from the solid phase to the liquid phase without changing its temperature. In our exercise, we deal with freezing water, so we are focusing on the inverse process.
For water, the latent heat of fusion is given as 333 kJ/kg. This means that 333 kJ of energy must be removed from 1 kg of liquid water at 0°C to convert it to ice at 0°C. This energy removal happens without a change in temperature, just like the heat added does not change the temperature during melting.
This concept is crucial in ice-making processes. By removing 333 kJ of energy per kg of water, we are able to convert it into ice. Understanding this helps in calculating the efficiency and effectiveness of thermal systems like ice makers.

Energy Conversion

Energy conversion is the process of changing energy from one form to another. In our example exercise, the ice maker converts electrical energy into thermal energy to remove heat from water, thus freezing it.
The power input we considered is 1 kW, equivalent to 1 kJ/s. To find out how much energy is applied over an hour, we multiply the power by the number of seconds in an hour (3600 seconds), yielding 3600 kJ/h.
This energy conversion process is vital in various household and industrial applications. Each application, like water freezing, requires a specific amount of energy conversion, based on the substance's physical properties, such as the latent heat of fusion.
Having a good grasp of these conversion principles allows efficient design and operation of thermal systems.

Steady State Processes

In thermodynamics, a steady state process is a condition where all state variables remain constant despite ongoing processes. In this context, our ice maker operates at a steady state, meaning the rate at which energy is added equals the rate at which energy is removed.
Because the ice maker is in steady state, the power input is consistently being converted to thermal energy to remove heat from water. This consistent removal of energy ensures that the rate of ice production remains constant.
Calculations in steady state make theoretical analysis simpler because we do not account for changes over time—everything is considered in equilibrium.
This allows the determination of maximum rates of processes, like ice production, under specified conditions. Understanding steady state helps in predicting the performance and efficiency of devices and systems across various applications.