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Solved: When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 106P Chapter 3

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 106P

When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a 12.0-oz drink from \(75\ ^{\circ}\mathrm{F}\) to \(35\ ^{\circ}\mathrm{F}\), if the heat capacity of the drink is \(4.18\mathrm{\ J}/\mathrm{g}^{\circ}\mathrm{C}\)? (Assume that the heat transfer is 100% efficient.)

Equation Transcription:

Text Transcription:

75 deg F

35 deg F

4.18 J/g deg C

Step-by-Step Solution:
Step 1 of 3

Solution 106P

In this question it has given that the heat capacity of the drink= 4.18 J/g °C

Heat absorbs (q) = 0.33 KJ per gram

T2= 75oF

T1 = 35oF

= T2 -T1 = (75-35)oF = 40oF

It is known that,

1.0 oz = 28.3g.

Hence 12.0 oz will be equal to,

12.0 oz x 28.35g/oz = 339.6g (m)

The heat absorbed (q) = mCΔT

           = (339.6 g) x (4.184 J/g C) x (40 )

                                    = 56.8 kJ 1 g 56.8 kJ x

           = 170 g of ice 0.33 kJ

Step 2 of 3

Chapter 3, Problem 106P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Solved: When ice melts, it absorbs 0.33 kJ per gram. How much ice is required to cool a