/*! 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 21 It takes \(880 \mathrm{J}\) to r... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 21

It takes \(880 \mathrm{J}\) to raise the temperature of \(350 \mathrm{g}\) of lead from 0 to \(20.0^{\circ} \mathrm{C} .\) What is the specific heat of lead?

Short Answer

Expert verified
Answer: The specific heat of lead is \(0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\).

Step by step solution

01

Identify the given values

We are given the following information: - Amount of energy required to raise the temperature (Q) = 880 J - Mass of lead (m) = 350 g - Initial temperature (T1) = 0掳C - Final temperature (T2) = 20.0掳C
02

Write down the formula for energy change

The formula for energy change, which relates energy, mass, specific heat, and temperature change is: \(Q = mc (T_{2} - T_{1})\) We need to solve for the specific heat of lead (c), so we can rewrite the formula as follows: \(c = \frac{Q}{m(T_{2} - T_{1})}\)
03

Plug in the given values into the formula

Now we can substitute the given values into the equation: \(c = \frac{880 \mathrm{J}}{350 \mathrm{g}(20.0^{\circ}\mathrm{C}-0^{\circ}\mathrm{C})}\)
04

Calculate the specific heat

Simplify and compute the value for the specific heat: \(c = \frac{880 \mathrm{J}}{350 \mathrm{g}(20.0^{\circ}\mathrm{C})}\) \(c = \frac{880 \mathrm{J}}{7000 \mathrm{g}\cdot^{\circ}\mathrm{C}}\) \(c = 0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\)
05

Write the final answer

The specific heat of lead is \(0.1257 \frac{\mathrm{J}}{\mathrm{g}\cdot^{\circ}\mathrm{C}}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

A piece of gold of mass \(0.250 \mathrm{kg}\) and at a temperature of \(75.0^{\circ} \mathrm{C}\) is placed into a \(1.500-\mathrm{kg}\) copper pot containing \(0.500 \mathrm{L}\) of water. The pot and water are at $22.0^{\circ} \mathrm{C}$ before the gold is added. What is the final temperature of the water?

Consider the leaf of Problem \(70 .\) Assume that the top surface of the leaf absorbs \(70.0 \%\) of \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2}\) of radiant energy, while the bottom surface absorbs all of the radiant energy incident on it due to its surroundings at \(25.0^{\circ} \mathrm{C} .\) (a) If the only method of heat loss for the leaf were thermal radiation, what would be the temperature of the leaf? (Assume that the leaf radiates like a blackbody.) (b) If the leaf is to remain at a temperature of \(25.0^{\circ} \mathrm{C}\) how much power per unit area must be lost by other methods such as transpiration (evaporative heat loss)?
A spring of force constant \(k=8.4 \times 10^{3} \mathrm{N} / \mathrm{m}\) is compressed by \(0.10 \mathrm{m} .\) It is placed into a vessel containing $1.0 \mathrm{kg}$ of water and then released. Assuming all the energy from the spring goes into heating the water, find the change in temperature of the water.
What is the heat capacity of a system consisting of (a) a \(0.450-\mathrm{kg}\) brass cup filled with \(0.050 \mathrm{kg}\) of water? (b) \(7.5 \mathrm{kg}\) of water in a 0.75 -kg aluminum bucket?
An incandescent lightbulb has a tungsten filament that is heated to a temperature of \(3.00 \times 10^{3} \mathrm{K}\) when an electric current passes through it. If the surface area of the filament is approximately $1.00 \times 10^{-4} \mathrm{m}^{2}\( and it has an emissivity of \)0.32,$ what is the power radiated by the bulb?
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Fundamentals of Physics

Read Explanation

Quantum Physics

Read Explanation

Oscillations

Read Explanation

Scientific Method Physics

Read Explanation

Atoms and Radioactivity

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.