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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 3 - Problem 106p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 3 - Problem 106p

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

ISBN: 9780321910295 34

## Solution for problem 106P Chapter 3

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

Step 3 of 3

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