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Chapter 17: Problem 94

A cylindrical copper rod \(37 \mathrm{cm}\) long and \(7.5 \mathrm{cm}\) in diameter is placed upright on a hot plate held at a constant temperature of \(120^{\circ} \mathrm{C}\), as indicated in Figure \(\mathrm{T}-3 \mathrm{l}\). A small depression on top of the rod holds a \(25-g\) ice cube at an initial temperature of \(0.0^{\circ} \mathrm{C} .\) How long does it take for the ice cube to melt? Assume there is no heat loss through the vertical surface of the rod, and that the thermal conductivity of copper is \(390 \mathrm{W} /\left(\mathrm{m} \cdot \mathrm{C}^{\circ}\right)\).

Short Answer

Expert verified
It takes about 1.82 seconds for the ice cube to melt.

Step by step solution

01

Determine the heat required to melt the ice

The heat required (\( q \) ) to melt the ice is given by\[ q = m \cdot L_f \]where \( m = 25g = 0.025 \text{ kg} \) is the mass of the ice cube and \( L_f = 334000 \text{ J/kg} \) is the latent heat of fusion for ice. Calculate \( q \).
02

Calculate the heat conduction rate through the rod

The rate of heat conduction (\( \dot{q} \) ) through the rod is given by Fourier鈥檚 Law:\[ \dot{q} = k \cdot A \cdot \frac{\Delta T}{L} \]where:- \( k = 390 \text{ W/m掳C} \) is the thermal conductivity of copper.- \( A = \pi \cdot (3.75 \text{ cm})^2 = 4.415 \times 10^{-3} \text{ m}^2 \) is the cross-sectional area of the rod.- \( \Delta T = 120掳C \) is the temperature difference across the rod.- \( L = 37 \text{ cm} = 0.37 \text{ m} \) is the length of the rod.Plug in the values and compute \( \dot{q} \).
03

Calculate the time to melt the ice

The time \( t \) required to melt the ice is given by:\[ t = \frac{q}{\dot{q}} \]where \( q \) is the heat required to melt the ice from Step 1, and \( \dot{q} \) is the heat conduction rate from Step 2. Calculate \( t \).

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a property that describes how well a material can conduct heat. It tells us how quickly thermal energy passes through a material. For good conductors like metals, this value is quite high. Copper, known for its excellent heat conduction capabilities, has a thermal conductivity of
390 W/(m掳C). This means that copper can transfer 390 joules of heat per second across a 1-meter cube of it with a 1掳C difference in temperature.
  • High thermal conductivity = efficient heat transfer
  • Copper's thermal conductivity is one of the reasons it's used in many heat-exchange applications, like cookware and heat sinks.
Thermal conductivity is crucial in calculating how much heat flows through a material over time. Understanding this helps engineers and scientists design systems where thermal energy conduction is important, like in the case of our copper rod scenario.
Latent Heat of Fusion
Latent heat of fusion is the amount of heat needed to change a substance from a solid to a liquid without changing its temperature. For ice, this value is
334,000 J/kg. This means that to turn 1 kilogram of ice at its melting point into water, 334,000 joules of energy is required.
  • Latent heat relates to phase changes, not temperature changes.
  • It explains why ice takes time to melt even when the temperature is above 0掳C.
For our exercise, melting a 25-gram ice cube requires multiplying the latent heat of fusion by the mass of the ice. This gives us a total energy value required to melt the entire cube, which helps in determining the time required when coupled with the heat conduction rate through the rod.
Heat Conduction Rate
The heat conduction rate, represented by \( \dot{q} \), describes how much heat flows through a material per unit of time. Fourier's Law of Heat Conduction gives a way to calculate this: \[ \dot{q} = k \cdot A \cdot \frac{\Delta T}{L} \]where all these factors:
  • \( k \) - thermal conductivity
  • \( A \) - cross-sectional area
  • \( \Delta T \) - temperature difference
  • \( L \) - length of the conducting path
contribute to how quickly heat will move through a material. In the context of our problem, knowing \( \dot{q} \) helps us find out how much heat is transferred to the ice over time, making it possible to calculate the time it takes for the ice to completely melt.

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Most popular questions from this chapter

A balloon contains 3.7 liters of nitrogen gas at a temperature of \(87 \mathrm{K}\) and a pressure of \(101 \mathrm{kPa}\). If the temperature of the gas is allowed to increase to \(24^{\circ} \mathrm{C}\) and the pressure remains constant, what volume will the gas occupy?

The air in your room is composed mostly of oxygen \(\left(\mathrm{O}_{2}\right)\) and nitrogen \(\left(\mathrm{N}_{2}\right)\) molecules. The oxygen molecules are more massive than the nitrogen molecules. (a) Is the rms speed of the \(\mathrm{O}_{2}\) molecules greater than, less than, or equal to the rms speed of the \(\mathrm{N}_{2}\) molecules? (b) Choose the best explanation from among the following: I. The more massive oxygen molecules have greater momentum and therefore greater speed. II. Equal temperatures for the oxygen and nitrogen molecules imply they have equal rms speeds. III. The temperature is the same for both molecules, and hence their average kinetic energies are equal. As a result, the more massive oxygen molecules have lower speeds.

A hollow cylindrical rod (rod 1) and a a solid cylindrical rod (rod 2 ) are made of the same material. The two rods have the same length and the same outer radius. If the same compressional force is applied to each rod, (a) is the change in length of rod 1 greater than, less than, or equal to the change in length of rod \(2 ?\) (b) Choose the best explanation from among the following: I. The solid rod has the larger effective cross-sectional area, since the empty part of the hollow rod doesn't resist compression. Therefore, the solid rod has the smaller change in length. II. The rods have the same outer radius and hence the same cross-sectional area. As a result, their change in length is the same. III. The walls of the hollow rod are hard and resist compression more than the uniform material in the solid rod. Therefore the hollow rod has the smaller change in length.

In the morning, when the temperature is \(286 \mathrm{K}\), a bicyclist finds that the absolute pressure in his tires is 501 kPa. That afternoon he finds that the pressure in the tires has increased to 554 kPa. lgnoring expansion of the tires, find the afternoon temperature.

As you go up in altitude, do you expect the ratio of oxygen to nitrogen in the atmosphere to increase, decrease; or stay the same? Explain.
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