/*! 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 58 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\(^\circ\)C, and the other end is maintained at 0.00\(^\circ\)C by an ice-water mixture. The rod is 60.0 cm long and has a cross-sectional area of 1.25 cm\(^2\). The heat conducted by the rod melts 8.50 g of ice in 10.0 min. Find the thermal conductivity \(k\) of the metal.

Short Answer

Expert verified
Thermal conductivity \(k\) is 22.71 W/m·K.

Step by step solution

01

Understand the Problem

We need to find the thermal conductivity \(k\) of the metal rod given the amount of ice melted by heat conduction through the rod over a specific time period. We have the rod's temperature difference, dimensions, and the mass of ice melted.
02

Use the Heat Conducted Formula

The formula for heat conducted through a rod is \(Q = k \cdot A \cdot \Delta T \cdot \frac{t}{L}\), where \(Q\) is the heat transferred, \(A\) is the cross-sectional area, \(\Delta T\) is the temperature difference, \(t\) is the time, and \(L\) is the length of the rod.
03

Calculate the Heat Required 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 of fusion of ice (334 J/g). For 8.50 g of ice, \(Q = 8.50 \, \text{g} \times 334 \, \text{J/g} = 2839 \, \text{J}\).
04

Relate to the Conducted Heat Formula

We substitute the known values into the heat conducted formula \(Q = k \cdot A \cdot \Delta T \cdot \frac{t}{L}\) and solve for \(k\). Given \(Q = 2839 \, \text{J}\), \(A = 1.25 \, \text{cm}^2 = 1.25 \times 10^{-4} \, \text{m}^2\), \(\Delta T = 100.0 \, ^\circ\text{C}\), \(t = 10.0 \, \text{min} = 600 \, \text{s}\), and \(L = 60.0 \, \text{cm} = 0.60 \, \text{m}\).
05

Solve for Thermal Conductivity \(k\)

Rearrange the formula to find \(k\):\[k = \frac{Q \cdot L}{A \cdot \Delta T \cdot t} = \frac{2839 \, \text{J} \times 0.60 \, \text{m}}{1.25 \times 10^{-4} \, \text{m}^2 \times 100.0 \, ^\circ\text{C} \times 600 \, \text{s}}.\]Calculate to find \(k\).
06

Perform the Calculation

Substitute the values into the equation:\[k = \frac{2839 \, \text{J} \times 0.60 \, \text{m}}{1.25 \times 10^{-4} \, \text{m}^2 \times 100.0 \, ^\circ\text{C} \times 600 \, \text{s}} = 22.71 \, \text{W/m} \cdot \text{K}.\]

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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 is exchanged between two systems or bodies at different temperatures. In the context of the metal rod exercise, heat is transferred from the warmer end at 100.0\(^\circ\)C to the colder end in contact with the ice-water mixture, which is maintained at 0.0\(^\circ\)C.

This transfer continues until thermal equilibrium is reached, or as long as the temperature difference is maintained. The metal rod acts as a conduit for heat flow, showcasing the principle that heat moves from higher to lower temperatures. In practical applications, understanding heat transfer helps us design efficient systems for maintaining desired temperatures in engineering and everyday scenarios.
  • Temperature differential is essential for heat transfer to occur.
  • Heat always travels from hot to cold areas.
  • The rate of transfer depends on material properties like thermal conductivity.
Latent Heat of Fusion
The latent heat of fusion is the amount of heat absorbed or released by a substance as it changes state from solid to liquid or vice versa at constant temperature. In our exercise with the metal rod, the latent heat of fusion relates to the energy required to melt the ice.

For ice, the latent heat of fusion is 334 J/g, meaning it takes 334 joules to melt one gram of ice without changing its temperature. The formula \(Q = m \cdot L_f\) calculates this energy needed, where \(m\) is the mass of ice and \(L_f\) is the latent heat of fusion.
  • This concept is critical in calculating energy transitions in phase changes.
  • The heat required does not cause a temperature change but changes ice to water.
  • Understanding latent heat allows for better energy management in systems.
Temperature Gradient
The temperature gradient is the rate of temperature change with respect to distance through a material. It is a central concept in calculating heat transfer efficiency because a higher gradient typically indicates a faster heat transfer.

In the context of the insulated metal rod, the temperature gradient exists between the high temperature at one end (100.0\(^\circ\)C) and the low temperature at the other end (0.0\(^\circ\)C). This creates a consistent drive for heat to flow along the rod.
  • The temperature gradient drives thermal energy movement.
  • Larger gradients can lead to faster heat conduction.
  • Gradients are essential in solving problems involving heat exchange.
Thermal Conductivity Calculation
Thermal conductivity is a measure of a material's ability to conduct heat. It is denoted by \(k\) and is crucial for determining how quickly heat can pass through a substance.

In our example, thermal conductivity calculation involves using the rearranged formula \(k = \frac{Q \cdot L}{A \cdot \Delta T \cdot t}\). Here, \(Q\) represents the heat transferred, \(L\) is the length of the rod, \(A\) is its cross-sectional area, \(\Delta T\) is the temperature difference, and \(t\) is the time duration.

Substituting the given values, the calculation results in \(k\) for the metal rod. This calculation is vital for designing and understanding systems where controlled heat transfer is necessary, such as in insulations and heat exchangers.
  • High \(k\) values indicate better conductor of heat.
  • The material's physical properties directly affect \(k\).
  • Calculated \(k\) helps optimize thermal system designs.

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Most popular questions from this chapter

A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0\(^\circ\)C, into a calorimeter. The calorimeter can, initially at 19.0\(^\circ\)C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can and contents is 26.1\(^\circ\)C. Compute the specific heat of the sample.

BIO Before going in for his annual physical, a 70.0-kg man whose body temperature is 37.0\(^\circ\)C consumes an entire 0.355-L can of a soft drink (mostly water) at 12.0\(^\circ\)C. (a) What will his body temperature be after equilibrium is attained? Ignore any heating by the man’s metabolism. The specific heat of the man’s body is 3480 J/kg \(\cdot\) K. (b) Is the change in his body temperature great enough to be measured by a medical thermometer?

During your mechanical engineering internship, you are given two uniform metal bars \(A\) and \(B\), which are made from different metals, to determine their thermal conductivities. Measuring the bars, you determine that both have length 40.0 cm and uniform cross-sectional area 2.50 cm\(^2\). You place one end of bar \(A\) in thermal contact with a very large vat of boiling water at 100.0\(^\circ\)C and the other end in thermal contact with an ice-water mixture at 0.0\(^\circ\)C. To prevent heat loss along the bar's sides, you wrap insulation around the bar. You weigh the amount of ice initially and find it to be 300 g. After 45.0 min has elapsed, you weigh the ice again and find that 191 g of ice remains. The ice-water mixture is in an insulated container, so the only heat entering or leaving it is the heat conducted by the metal bar. You are confident that your data will allow you to calculate the thermal conductivity \(k_A\) of bar \(A\). But this measurement was tedious-you don't want to repeat it for bar \(B\). Instead, you glue the bars together end to end, with adhesive that has very large thermal conductivity, to make a composite bar 80.0 m long. You place the free end of A in thermal contact with the boiling water and the free end of \(B\) in thermal contact with the ice-water mixture. As in the first measurement, the composite bar is thermally insulated. You go to lunch; when you return, you notice that ice remains in the ice-water mixture. Measuring the temperature at the junction of the two bars, you find that it is 62.4\(^\circ\)C. After 10 minutes you repeat that measurement and get the same temperature, with ice remaining in the ice-water mixture. From your data, calculate the thermal conductivities of bar \(A\) and of bar \(B\).

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m\(^2\). The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

If the air temperature is the same as the temperature of your skin (about 30\(^\circ\)C), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical 70-kg person's body produces energy at a rate of about 500 W due to metabolism, 80% of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many 750-mL bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is 1.0 kg.)

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