/*! 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 56 One end of an insulated metal ro... [FREE SOLUTION] | 91Ó°ÊÓ

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One end of an insulated metal rod is maintained at \(100.0^{\circ} \mathrm{C}\), and the other end is maintained at \(0.00^{\circ} \mathrm{C}\) by an ice-water mixture. The rod is \(60.0 \mathrm{~cm}\) long and has a cross-sectional area of \(1.25 \mathrm{~cm}^{2}\). The heat conducted by the rod melts \(8.50 \mathrm{~g}\) of ice in \(10.0 \mathrm{~min} .\) Find the thermal conductivity \(k\) of the metal.

Short Answer

Expert verified
The thermal conductivity of the metal is \( k = 227.0 \, W/(m \cdot K) \).

Step by step solution

01

Determine the Heat Transfer

First, determine the heat transferred. The heat transferred is used to melt the ice. The heat required to melt ice is given by \(Q = m \cdot L_f\), where \(m\) is the mass of the ice and \(L_f\) is the latent heat fusion of ice. The mass of the ice is \(m = 8.5 g = 0.0085 kg\), and the latent heat fusion of ice is \(L_f = 334 \, kJ/kg = 334000 \, J/kg\). So the total heat transferred is \(Q = 0.0085kg \cdot 334000 J/kg = 2839 J\).
02

Calculate the Time

Then, determine the time during which the heat transfer occurred. It is given as 10 min, which should be transformed to the SI unit (seconds). So, the time is \(t = 10 min = 600s\). The heat rate \(q\) can then be obtained by dividing the total heat by the time i.e., \( q = Q/t = 2839J / 600s = 4.732 J/s \).
03

Determine the Temperature Difference

Next, identify the temperature difference across the rod. Here, one end of the rod is maintained at \(100^{\circ} \mathrm{C}\), while the other end is maintained at \(0^{\circ} \mathrm{C}\). Therefore, the temperature difference is \(\Delta T = (100 - 0)^{\circ} \mathrm{C} = 100^{\circ} \mathrm{C}\).
04

Calculate the Thermal Conductivity

Finally, calculate the thermal conductivity, \(k\), using the formula for heat conduction. Rearrange the formula to solve for \(k\): \( k = q \cdot d / (A \cdot \Delta T) \). The given values in SI units are the cross-sectional area \(A = 1.25 \, cm^{2} = 1.25 \times 10^{-4} \, m^{2}\), the distance \(d = 60.0 \, cm = 0.6 \, m\), and the heat rate \(q = 4.732 \, J/s\). Substitute these values in to find: \( k = 4.732 J/s \cdot 0.6m / (1.25 \times 10^{-4}m^{2} \cdot 100^{\circ} \mathrm{C}) = 227.0 \, W/(m \cdot K)\).

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Key Concepts

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

Heat Transfer
Understanding heat transfer is crucial when studying thermal conductivity and how energy moves in materials. Heat transfer refers to the movement of heat from one place to another, driven by a temperature difference. Three main mechanisms of heat transfer exist: conduction, convection, and radiation. In the context of our exercise, we're particularly interested in conduction, which occurs when heat energy travels through a material without the material itself moving. This is the process we see in the metal rod within the exercise.

For effective learning, imagine the metal rod as a busy highway where thermal energy is the cars speeding from the hot end to the cold end. The thermal conductivity of the material is like the road's capacity to accommodate traffic; higher conductivity means the road allows more 'thermal cars' to pass through effortlessly. The exercise asked to calculate the thermal conductivity (k) from the amount of ice melted, implying the real-world application of these principles – how a substance's ability to conduct heat can result in physical changes in another substance, such as turning ice into water.
Latent Heat Fusion
Latent heat fusion is another fascinating concept that often confuses students. It refers to the energy required to change a substance's state from solid to liquid at a constant temperature. It's called 'latent' because during the phase change, the energy is 'hidden' and does not cause a temperature increase. In our exercise, the ice absorbs latent heat as it melts, which doesn't raise the temperature but instead changes the structure of the water molecules.

To visualize this, think of a block of ice as a tightly packed dance floor. When latent heat is added, the dancers (water molecules) have enough energy to break apart and move more freely, akin to switching from a slow dance to a more lively one. They're still at the same spot - the temperature hasn't increased, but the state has now changed from solid ice to liquid water. This part was integral in solving our exercise, since we calculated the heat (energy) transferred to melt the ice using the mass of the ice and the latent heat of fusion. It provided the 'Q' in the subsequent steps needed to find the metal’s thermal conductivity.
Temperature Difference
The last but certainly not the least core concept is the temperature difference, often denoted as ΔT. It's a straightforward idea that the heat transfer rate between two regions in a material is proportional to the temperature gradient or difference between these regions. High school students often struggle with the distinction between heat and temperature; temperature measures the average kinetic energy of particles in a substance, while heat is the total energy transferred.

In the case of our insulated metal rod, where one end is kept at boiling temperature and the other at freezing point, this temperature difference is the driving force behind the conductive heat transfer, effectively 'pushing' the heat from hot to cold. The rod acts like an energy conveyor belt, moving heat from the high-energy (hot) side to the low-energy (cold) side. This is crucial in calculating the thermal conductivity (k) because it determines how effectively the material conducts the heat, which in practical terms, is central to applications like insulation in buildings and designing heat exchangers.

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Most popular questions from this chapter

A copper sphere with density \(8900 \mathrm{~kg} / \mathrm{m}^{3},\) radius \(5.00 \mathrm{~cm}\) and emissivity \(e=1.00\) sits on an insulated stand. The initial temperature of the sphere is \(300 \mathrm{~K}\). The surroundings are very cold, so the rate of absorption of heat by the sphere can be neglected. (a) How long does it take the sphere to cool by \(1.00 \mathrm{~K}\) due to its radiation of heat energy? Neglect the change in heat current as the temperature decreases. (b) To assess the accuracy of the approximation used in part (a), what is the fractional change in the heat current \(H\) when the temperature changes from \(300 \mathrm{~K}\) to \(299 \mathrm{~K} ?\)

An insulated beaker with negligible mass contains \(0.250 \mathrm{~kg}\) of water at \(75.0^{\circ} \mathrm{C}\). How many kilograms of ice at \(-20.0^{\circ} \mathrm{C}\) must be dropped into the water to make the final temperature of the system \(40.0^{\circ} \mathrm{C}\) ?

In a container of negligible mass, \(0.0400 \mathrm{~kg}\) of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to \(0.200 \mathrm{~kg}\) of water at \(50.0^{\circ} \mathrm{C}\) (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

An electric kitchen range has a total wall area of \(1.40 \mathrm{~m}^{2}\) and is insulated with a layer of fiberglass \(4.00 \mathrm{~cm}\) thick. The inside surface of the fiberglass has a temperature of \(175^{\circ} \mathrm{C},\) and its outside surface is at \(35.0^{\circ} \mathrm{C}\). The fiberglass has a thermal conductivity of \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of \(1.40 \mathrm{~m}^{2}\) ? (b) What electric-power input to the heating element is required to maintain this temperature?

In a container of negligible mass, \(0.200 \mathrm{~kg}\) of ice at an initial temperature of \(-40.0^{\circ} \mathrm{C}\) is mixed with a mass \(m\) of water that has an initial temperature of \(80.0^{\circ} \mathrm{C}\). No heat is lost to the surroundings. If the final temperature of the system is \(28.0^{\circ} \mathrm{C},\) what is the mass \(m\) of the water that was initially at \(80.0^{\circ} \mathrm{C}\) ?

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