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The specific heat capacity of ice is about 0.5 cal/g°C.

Conceptual Physics | 12th Edition | ISBN: 9780321909107 | Authors: Paul G. Hewitt ISBN: 9780321909107 29

Solution for problem 2P Chapter 17

Conceptual Physics | 12th Edition

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Conceptual Physics | 12th Edition | ISBN: 9780321909107 | Authors: Paul G. Hewitt

Conceptual Physics | 12th Edition

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Problem 2P

The specific heat capacity of ice is about 0.5 cal/g°C. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1-g ice cube at absolute zero (?273°C) to 1 g of boiling water. How does this number of calories compare with the number of calories required to change the same gram of 100°C boiling water to 100°C steam?

Step-by-Step Solution:

Solution 2P Introductio The heat required to change ice from absolute zero to its boiling point can be divided in three steps. First heat required to change the temperature from absolute zero to melting point. Then the latent heat required to change its phase from solid to liquid and in final step, heat required to change temperature of the water from melting point to boiling point. Solution Part I Step 1 The heat required to change temperature is given by Q = mcT Now to change the temperature of the ice from absolute zero (-273°C) to melting point (0°C) is given by Q 1 mc ice ice Here we have m = 1 g is the mass of the ice, ice= 0.5 cal/g°C is the specific heat and T = (0°C ( 273°C) = 273°C is the change in temperature of the ice, therefore we have ice Q 1 (1 g)(0.5 cal/g°C)(273°C) = 136.5 cal Step 2 The heat required to melt ice is given by Q = mL 2 Where m = 1 g is the mass of the ice, L = 80 cal/g is the latent heat of the ice. Hence the required heat is Q = (1 g)(80 cal/g) = 80 cal 2 Step 3 Now, we have to supply heat to change the temperature of the water from 0°C to 100°C . The required heat is given by Q = mc T 3 water water Here we have m = 1 g is the mass of the water, c water= 1 cal/g°C is the specific heat and T = (100°C 0°) = 100°C is the change in temperature of the water. Hence, we have water Q 2 (1 g)(1 cal/g°C)(100°C) = 100 cal

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Chapter 17, Problem 2P is Solved
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Textbook: Conceptual Physics
Edition: 12
Author: Paul G. Hewitt
ISBN: 9780321909107

Since the solution to 2P from 17 chapter was answered, more than 311 students have viewed the full step-by-step answer. The answer to “The specific heat capacity of ice is about 0.5 cal/g°C. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1-g ice cube at absolute zero (?273°C) to 1 g of boiling water. How does this number of calories compare with the number of calories required to change the same gram of 100°C boiling water to 100°C steam?” is broken down into a number of easy to follow steps, and 72 words. The full step-by-step solution to problem: 2P from chapter: 17 was answered by , our top Physics solution expert on 04/03/17, 08:01AM. This full solution covers the following key subjects: Calories, ice, Water, boiling, absolute. This expansive textbook survival guide covers 45 chapters, and 4650 solutions. This textbook survival guide was created for the textbook: Conceptual Physics, edition: 12. Conceptual Physics was written by and is associated to the ISBN: 9780321909107.

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