Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 48

A metal rod with a diameter of \(2.30 \mathrm{cm}\) and length of $1.10 \mathrm{m}\( has one end immersed in ice at \)32.0^{\circ} \mathrm{F}$ and the other end in boiling water at \(212^{\circ} \mathrm{F}\). If the ice melts at a rate of 1.32 g every 175 s, what is the thermal conductivity of this metal? Identify the metal. Assume there is no heat lost to the surrounding air.

Short Answer

Expert verified
Answer: The thermal conductivity of the metal rod is 0.67 J/(cm s °C), and it is most likely made of iron.

Step by step solution

01

Convert Fahrenheit temperatures to Celsius

First, we need to convert the Fahrenheit temperatures to Celsius. This is because the units of thermal conductivity, which we will be calculating, typically depend on the Celsius scale. To convert from Fahrenheit to Celsius, use the formula: Celsius = (Fahrenheit - 32) * 5/9 For the ice: Celsius_ice = (32 - 32) * 5/9 = 0°C For the boiling water: Celsius_water = (212 - 32) * 5/9 = 100°C
02

Determine potential energy required to melt ice

Now we will determine the amount of potential energy Q required to melt the ice. Use the formula: Q = mass × specific_heat_of_ice × temperature_difference Where mass is given in the problem statement as 1.32 g and the specific heat of ice is 334 J/g. Potential_energy_to_melt_ice = 1.32 g × 334 J/g = 440.88 J
03

Determine rate of heat flow

The heat flow rate, which represents the amount of heat transferred per unit of time, can be calculated using the formula below: Heat_flow_rate = Potential_energy_to_melt_ice / time Where time is given as 175 seconds. Heat_flow_rate = 440.88 J / 175 s = 2.52 J/s
04

Calculate thermal conductivity

Now we will use the formula for heat conduction through a cylindrical rod: Heat_flow_rate = (thermal_conductivity × area × temperature_difference) / length Solving for thermal_conductivity with the information given in the problem: thermal_conductivity = (Heat_flow_rate × length) / (area × temperature_difference) First, we need to calculate the area of the rod. The area of a cylinder can be found using the formula: Area = π × (diameter / 2)^2 Area = π × (2.3 cm / 2)^2 = 4.15 cm² We also need to convert the length to centimeters: Length = 1.1 m × 100 cm/m = 110 cm Now we can plug in all the values we obtained so far: thermal_conductivity = (2.52 J/s × 110 cm) / (4.15 cm² × 100°C) = 0.67 J/(cm s °C)
05

Identify the metal

Now we can compare the thermal conductivity value we found to known thermal conductivity values of different metals: Aluminum: 2.37 J/(cm s °C) Copper: 3.93 J/(cm s °C) Brass: 1.06 J/(cm s °C) Iron: 0.67 J/(cm s °C) etc. The result we obtained is closest to the value for iron, suggesting that the metal rod is made of iron.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An incandescent light bulb radiates at a rate of \(60.0 \mathrm{W}\) when the temperature of its filament is \(2820 \mathrm{K}\). During a brownout (temporary drop in line voltage), the power radiated drops to \(58.0 \mathrm{W} .\) What is the temperature of the filament? Neglect changes in the filament's length and cross-sectional area due to the temperature change. (tutorial: light bulb)
What mass of water at \(25.0^{\circ} \mathrm{C}\) added to a Styrofoam cup containing two 50.0 -g ice cubes from a freezer at \(-15.0^{\circ} \mathrm{C}\) will result in a final temperature of \(5.0^{\circ} \mathrm{C}\) for the drink?
A child of mass \(15 \mathrm{kg}\) climbs to the top of a slide that is $1.7 \mathrm{m}\( above a horizontal run that extends for \)0.50 \mathrm{m}$ at the base of the slide. After sliding down, the child comes to rest just before reaching the very end of the horizontal portion of the slide. (a) How much internal energy was generated during this process? (b) Where did the generated energy go? (To the slide, to the child, to the air, or to all three?)
It requires \(17.10 \mathrm{kJ}\) to melt \(1.00 \times 10^{2} \mathrm{g}\) of urethane $\left[\mathrm{CO}_{2}\left(\mathrm{NH}_{2}\right) \mathrm{C}_{2} \mathrm{H}_{5}\right]\( at \)48.7^{\circ} \mathrm{C} .$ What is the latent heat of fusion of urethane in \(\mathrm{kJ} / \mathrm{mol} ?\)
A 2.0 -kg block of copper at \(100.0^{\circ} \mathrm{C}\) is placed into $1.0 \mathrm{kg}$ of water in a 2.0 -kg iron pot. The water and the iron pot are at \(25.0^{\circ} \mathrm{C}\) just before the copper block is placed into the pot. What is the final temperature of the water, assuming negligible heat flow to the environment?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Scientific Method Physics

Read Explanation

Oscillations

Read Explanation

Linear Momentum

Read Explanation

Radiation

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free