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Chapter 15: Problem 56

The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The maximum amount of ice that can be produced in 2 hours is approximately 30.4 kg.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin to work with them in thermodynamic calculations. To do this, we will add 273.15 to each temperature. \(T_H = 20.0^\circ C + 273.15 K = 293.15 K\) \(T_C = -5.0^\circ C + 273.15 K = 268.15 K\) \(T_W = 8.0^\circ C + 273.15 K = 281.15 K\)
02

Calculate the Coefficient of performance (COP) for the refrigerator

For a reversible refrigerator, the Coefficient of Performance (COP) is given by the following formula: \(\text{COP} = \frac{T_C}{T_H-T_C}\) Plug in the values of \(T_H\) and \(T_C\) to get the COP. \(\text{COP} = \frac{268.15}{293.15 - 268.15} = \frac{268.15}{25} = 10.726\)
03

Determine the heat removed from the cold reservoir

Since we know the power output of the refrigerator (148 W), we can calculate the heat removed from the cold reservoir using the formula: \(Q_C = \text{COP} \times W = 10.726 \times 148 W = 1587.448 W\) Now we need to find the heat removed in 2 hours, so we multiply this by the time: \(Q_C = 1587.448 W \times 2 h \times 3600 s/h = 11445504.64 J\)
04

Calculate the heat removed from the water to produce ice

To produce ice from the water, firstly, we need to cool down the water from \(281.15 K\) to \(273.15 K\), and then freeze it. The specific heat capacity of water (\(c_w\)) is \(4.18 \times 10^3 \mathrm{J/kg K}\) and the heat of fusion for water (\(L\)) is \(3.34 \times 10^5 \mathrm{J/kg}\). Let's denote the mass of water to be produced into ice by \(m\). Then, we have: \(\text{Heat removed to cool down water} = m \cdot c_w \cdot (T_W - 273.15 K )\) \(\text{Heat removed to freeze water} = m \cdot L \) The total heat removed can be found by summing these two terms, thus: \(Q_C = m \cdot (c_w \cdot 8 K + L)\)
05

Calculate the mass of ice produced

Now we can find the mass of ice produced by solving for \(m\): \(m = \frac{Q_C}{c_w \cdot 8 K + L} = \frac{11445504.64 J}{(4.18 \times 10^3 J/kgK) \cdot 8 K + 3.34 \times 10^5 J/kg} = 30.4 kg\) So, the maximum amount of ice that can be produced in 2 hours is approximately 30.4 kg.

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Most popular questions from this chapter

A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)
The United States generates about \(5.0 \times 10^{16} \mathrm{J}\) of electric energy a day. This energy is equivalent to work, since it can be converted into work with almost \(100 \%\) efficiency by an electric motor. (a) If this energy is generated by power plants with an average efficiency of \(0.30,\) how much heat is dumped into the environment each day? (b) How much water would be required to absorb this heat if the water temperature is not to increase more than \(2.0^{\circ} \mathrm{C} ?\)
Calculate the maximum possible efficiency of a heat engine that uses surface lake water at \(18.0^{\circ} \mathrm{C}\) as a source of heat and rejects waste heat to the water \(0.100 \mathrm{km}\) below the surface where the temperature is \(4.0^{\circ} \mathrm{C}\).
Two engines operate between the same two temperatures of \(750 \mathrm{K}\) and \(350 \mathrm{K},\) and have the same rate of heat input. One of the engines is a reversible engine with a power output of \(2.3 \times 10^{4} \mathrm{W} .\) The second engine has an efficiency of \(42 \% .\) What is the power output of the second engine?
An engincer designs a ship that gets its power in the following way: The engine draws in warm water from the ocean, and after extracting some of the water's internal energy, returns the water to the ocean at a temperature \(14.5^{\circ} \mathrm{C}\) lower than the ocean temperature. If the ocean is at a uniform temperature of \(17^{\circ} \mathrm{C},\) is this an efficient engine? Will the engineer's design work?
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