/*! 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 34 How much heat is required to cha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 34

How much heat is required to change \(1.0 \mathrm{kg}\) of ice, originally at \(-20.0^{\circ} \mathrm{C},\) into steam at \(110.0^{\circ} \mathrm{C} ?\) Assume 1.0 atm of pressure.

Short Answer

Expert verified
Answer: 3074.0 kJ

Step by step solution

01

Heating the ice from -20掳C to 0掳C

To find the heat needed in this step, we'll use the formula: \(Q = mc\Delta T\) - m: mass of the ice (1.0 kg) - c: specific heat capacity of ice (\(2.1 \frac{kJ}{kg\cdot 掳C}\)) - \(\Delta T\): temperature difference (-20掳C to 0掳C, so \(\Delta T = 20掳C\)) \(Q_1 = (1.0kg)(2.1\frac{kJ}{kg\cdot 掳C})(20 掳C) = 42.0 kJ\)
02

Melting the ice into water at 0掳C

To find the heat needed in this step, use the formula: \(Q = mL_f\) - m: mass of the ice (1.0 kg) - \(L_f\): latent heat of fusion for water (\(334 \frac{kJ}{kg}\)) \(Q_2 = (1.0kg)(334\frac{kJ}{kg}) = 334.0 kJ\)
03

Heating the water from 0掳C to 100掳C

To find the heat needed in this step, use the formula: \(Q = mc\Delta T\) - m: mass of the water (1.0 kg) - c: specific heat capacity of water (\(4.18 \frac{kJ}{kg\cdot 掳C}\)) - \(\Delta T\): temperature difference (0掳C to 100掳C, so \(\Delta T = 100掳C\)) \(Q_3 = (1.0kg)(4.18\frac{kJ}{kg\cdot 掳C})(100 掳C) = 418.0 kJ\)
04

Turning the water to steam at 100掳C

To find the heat needed in this step, use the formula: \(Q = mL_v\) - m: mass of the water (1.0 kg) - \(L_v\): latent heat of vaporization for water (\(2260 \frac{kJ}{kg}\)) \(Q_4 = (1.0kg)(2260\frac{kJ}{kg}) = 2260.0 kJ\)
05

Heating the steam from 100掳C to 110掳C

To find the heat needed in this step, use the formula: \(Q = mc\Delta T\) - m: mass of the steam (1.0 kg) - c: specific heat capacity of steam (\(2.0 \frac{kJ}{kg\cdot 掳C}\)) - \(\Delta T\): temperature difference (100掳C to 110掳C, so \(\Delta T = 10掳C\)) \(Q_5 = (1.0kg)(2.0\frac{kJ}{kg\cdot 掳C})(10 掳C) = 20.0 kJ\)
06

Summing the heat involved in all phases

Now we'll add the heat involved in all the steps to get the total heat required. \(Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 42.0 kJ + 334.0 kJ + 418.0 kJ + 2260.0 kJ + 20.0 kJ = 3074.0 kJ\) So, the amount of heat required to change 1.0 kg of ice originally at -20掳C into steam at 110掳C is 3074.0 kJ.

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 person of surface area \(1.80 \mathrm{m}^{2}\) is lying out in the sunlight to get a tan. If the intensity of the incident sunlight is $7.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},$ at what rate must heat be lost by the person in order to maintain a constant body temperature? (Assume the effective area of skin exposed to the Sun is \(42 \%\) of the total surface area, \(57 \%\) of the incident radiation is absorbed, and that internal metabolic processes contribute another \(90 \mathrm{W}\) for an inactive person.)
In a physics lab, a student accidentally drops a \(25.0-\mathrm{g}\) brass washer into an open dewar of liquid nitrogen at 77.2 K. How much liquid nitrogen boils away as the washer cools from \(293 \mathrm{K}\) to $77.2 \mathrm{K} ?\( The latent heat of vaporization for nitrogen is \)199.1 \mathrm{kJ} / \mathrm{kg}$.
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)
The thermal resistance of a seal's fur and blubber combined is $0.33 \mathrm{K} / \mathrm{W} .\( If the seal's internal temperature is \)37^{\circ} \mathrm{C}\( and the temperature of the sea is about \)0^{\circ} \mathrm{C},$ what must be the heat output of the seal in order for it to maintain its internal temperature?
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?
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Fields in Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electrostatics

Read Explanation

Quantum Physics

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.