   # Three 45-g ice cubes at 0 °C are dropped into 5.00 × 10 2 mL of tea to make iced tea. The tea was initially at 20.0 °C; when thermal equilibrium was reached, the final temperature was 0 °C. How much of the ice melted, and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water. ### 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, Problem 78GQ
Textbook Problem
490 views

## Three 45-g ice cubes at 0 °C are dropped into 5.00 × 102 mL of tea to make iced tea. The tea was initially at 20.0 °C; when thermal equilibrium was reached, the final temperature was 0 °C. How much of the ice melted, and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.

Interpretation Introduction

Interpretation:

The mass required for melting ice at definite volume at two different temperatures and remained floating beverage has to be calculated.

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 above equation to find out the sum of energy quantities.

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

Given,

The specific heat capacity is 4.2J/gK

Mass of the tea is 5×102g

The change in temperature is 20K

Substitute for the two energy quantities

4

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