   Chapter 5, Problem 77GQ

Chapter
Section
Textbook Problem

You add 100.0 g of water at 60.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the ice and water mixture reaches thermal equilibrium at 0 °C, how much ice has melted?

Interpretation Introduction

Interpretation:

At thermal equilibrium the amount ice melted 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.

The heat of the substance can be calculated as:

q=heat of fusion or vaporization × mass

Explanation

Given,

Mass = 100g

ΔT=-60K

The heat of the substance can be calculated as:

q=heat of fusion or vaporization × mass and q=C×m×ΔT

So, both the equation are compared

Let the amount of ice melted=x

Assume qwater=-qice

Specific heat capacity of water is 4

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