   # To make a glass of iced tea, you pour 250 mL of tea, whose temperature is 18.2 °C, into a glass containing five ice cubes. Each cube has a mass of 15 g. What quantity of ice will melt, and how much ice will remain floating in the tea? Assume iced tea has a density of 1.0 g/mL and a specific heat capacity of 4.2 J/g · K, that energy is transferred only as heat within the system, ice is at 0.0 °C, and no energy is transferred between system and surroundings. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 5.3, Problem 2CYU
Textbook Problem
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## To make a glass of iced tea, you pour 250 mL of tea, whose temperature is 18.2 °C, into a glass containing five ice cubes. Each cube has a mass of 15 g. What quantity of ice will melt, and how much ice will remain floating in the tea? Assume iced tea has a density of 1.0 g/mL and a specific heat capacity of 4.2 J/g · K, that energy is transferred only as heat within the system, ice is at 0.0 °C, and no energy is transferred between system and surroundings.

Interpretation Introduction

Interpretation:

The mass required for melting of ice at definite volume at two different temperatures and the amount of ice float on the tea has to be determined.

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1k.

Energy gained or lost can be calculated using the below equation.

q=C×m×ΔT

Where,

q= energy gained or lost for a given mass of substance (m)

C =specific heat capacity,

ΔT= change in temperature.

### Explanation of Solution

Assume the sum of two energy quantities are zero.

Use the below equation to find out the sum of energy quantities.

[Ctea×Mtea(Tfinal-Tinitial)+Cice×Mice(Tfinal-Tinitial)]=0

Substitute for the two energy quantities

Specific heat capacity of tea is 4

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