/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 To help prevent frost damage, fr... [FREE SOLUTION] | 91Ó°ÊÓ

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To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below freezing. When the water turns to ice during the night, heat is released into the plants, thereby giving a measure of protection against the cold. Suppose a grower sprays \(7.2 \mathrm{kg}\) of water at \(0^{\circ} \mathrm{C}\) onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a \(180-\mathrm{kg}\) tree rise if it absorbed the heat released in part (a)? Assume that the specific heat capacity of the tree is \(2.5 \times 10^{3} \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and that no phase change occurs within the tree itself.

Short Answer

Expert verified
The heat released is 2,404,800 J, and the tree's temperature rises by 5.34°C.

Step by step solution

01

Understanding the Problem

We need to find the heat released when 7.2 kg of water freezes at \(0^{\circ} \mathrm{C}\). Then, we need to calculate how this heat would affect the temperature of a 180 kg tree.
02

Calculate Heat Released from Freezing Water

When water freezes, it releases energy equal to the latent heat of fusion. The latent heat of fusion for water is \(334,000 \, \text{J/kg}\). To find the total heat released \(Q\) by 7.2 kg of water, use the formula:\[ Q = mL \]Where \(m\) is the mass of water and \(L\) is the latent heat of fusion.\[ Q = 7.2 \, \text{kg} \times 334,000 \, \text{J/kg} = 2,404,800 \, \text{J} \]
03

Calculate Temperature Rise of the Tree

Now, we need to find out by how much the temperature \(\Delta T\) of the 180 kg tree would rise if it absorbed all the heat released by the water. Use the formula:\[ Q = mc \Delta T \]rearranged as:\[ \Delta T = \frac{Q}{mc} \]Where \(m\) is the mass of the tree and \(c\) is the specific heat capacity of the tree \(2.5 \times 10^{3} \, \text{J/kg} \cdot \text{C}^{\circ}\).\[ \Delta T = \frac{2,404,800 \, \text{J}}{180 \, \text{kg} \times 2,500 \, \text{J/kg} \cdot \text{C}^{\circ}} = 5.34^{\circ} \text{C} \]
04

Review the Results

The freezing of 7.2 kg of water releases 2,404,800 J of heat. If this heat is absorbed by a 180 kg tree, its temperature will rise by approximately 5.34°C.

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Key Concepts

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

Heat Transfer
Heat transfer is the process through which thermal energy moves from one entity or mass to another. This transfer can occur in three main ways: conduction, convection, and radiation. In this particular exercise, we are dealing with a scenario of conduction—where heat energy is transferred directly from the warm ice (as it freezes) to the tree, thus warming it up.
When water freezes, it releases heat called the latent heat of fusion. This released energy helps to warm the surrounding area—in our case, a fruit tree's branches and leaves.
  • Conduction: Direct transfer of heat through a medium or direct contact.
  • Convection: Transfer of heat by the physical movement of a fluid (liquid or gas).
  • Radiation: Transfer of heat through electromagnetic waves, no medium required.
The main role of heat transfer in this problem is to mitigate the risk of frost damage by relying on the energy released during the freezing process. This energy is conducted to the tree, raising its temperature even during cold nights.
Specific Heat Capacity
Specific heat capacity is a measure of how much heat energy is needed to raise the temperature of a unit mass of a substance by 1 degree Celsius. It’s crucial in determining how substances respond to heat energy changes. In this scenario, the specific heat capacity of the tree plays a key role in calculating the consequent temperature change after it absorbs heat.
The tree’s specific heat capacity is given as \(2.5 imes 10^3 \, \text{J/kg} \cdot \text{C}^{\circ}\), which tells us how much energy is needed to change its temperature.
  • High specific heat capacity: Requires more energy to increase temperature, holds heat longer.
  • Low specific heat capacity: Heats up quickly with less energy, does not retain heat as well.
A higher specific heat capacity indicates a substance can absorb more heat without a significant rise in temperature, providing a slow and steady increase which is often beneficial for plants.
Temperature Change
Temperature change refers to how much the temperature of an object fluctuates based on heat absorption or release. In the case discussed, the temperature of the 180 kg tree changes because it absorbs the heat released by the freezing water.
The formula used to calculate the temperature change is \( \Delta T = \frac{Q}{mc} \). Here, \(Q\) is the heat absorbed from the water, \(m\) is the mass of the tree, and \(c\) is the specific heat capacity of the tree.
  • Temperature increase: Occurs when heat is absorbed by the system.
  • Temperature decrease: Results when the system loses heat.
This calculation illustrates how a tree can naturally raise its temperature from the heat gained, providing insight into how fruit growers utilize this natural phenomenon to save crops from frost damage.
Freezing Process
The freezing process is a fascinating phase change where a liquid turns into a solid. In this case, it’s the water transforming into ice which plays a dual role in plant protection against frost. The process releases latent heat, a crucial energy transition that can prevent crops from experiencing damaging frost temperatures.
Latent heat of fusion is the specific energy required for water to transition from a liquid to solid form without changing its temperature, releasing energy that can be absorbed by nearby materials, like trees.
  • Initiation of freezing: Water starts to solidify at 0°C but maintains this temperature while releasing latent heat.
  • Heat release: As each kilogram of water freezes, it releases approximately 334,000 J of energy.
Understanding the freezing process provides growers with a natural method to protect their crops by exploiting the energy changes involved and ensuring that the surrounding air and plants remain slightly warmer during freeze events.

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Most popular questions from this chapter

Occasionally, huge icebergs are found floating on the ocean's currents. Suppose one such iceberg is \(120 \mathrm{km}\) long, \(35 \mathrm{km}\) wide, and \(230 \mathrm{m}\) thick. (a) How much heat would be required to melt this iceberg (assumed to be at \(0^{\circ} \mathrm{C}\) ) into liquid water at \(0^{\circ} \mathrm{C}\) ? The density of ice is \(917 \mathrm{kg} / \mathrm{m}^{3}\). (b) The annual energy consumption by the United States is about \(1.1 \times 10^{20}\) J. If this energy were delivered to the iceberg every year, how many years would it take before the ice melted?

A thick, vertical iron pipe has an inner diameter of \(0.065 \mathrm{m}\). A thin aluminum disk, heated to a temperature of \(85^{\circ} \mathrm{C},\) has a diameter that is \(3.9 \times 10^{-5} \mathrm{m}\) greater than the pipe's inner diameter. The disk is laid on top of the open upper end of the pipe, perfectly centered on it, and allowed to cool. What is the temperature of the aluminum disk when the disk falls into the pipe? Ignore the temperature change of the pipe.

The latent heat of vaporization of \(\mathrm{H}_{2} \mathrm{O}\) at body temperature \(\left(37.0^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) To cool the body of a \(75-\mathrm{kg}\) jogger [average specific heat capacity \(\left.=3500 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\right]\) by \(1.5 \mathrm{C}^{\circ},\) how many kilograms of water in the form of sweat have to be evaporated?

An unknown material has a normal melting/freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) One-tenth of a kilogram of the solid at \(-25.0{ }^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains \(0.100 \mathrm{kg}\) of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C} .\) All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C} .\) The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

A piece of glass has a temperature of \(83.0^{\circ} \mathrm{C} .\) Liquid that has a temperature of \(43.0^{\circ} \mathrm{C}\) is poured over the glass, completely covering it, and the temperature at equilibrium is \(53.0^{\circ} \mathrm{C} .\) The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

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