Explanation Heat absorbed by ice at 0.0 0 C = Heat required to melt ice at 0.0 0 C + heat required to heat up water from 0.0 0 C to 24.0 0 C. Heat required to heat up water from 0.0 0 C to 24.0 0 C can be calculated using the following heat equation, Writing the heat equation and necessary conversion factors: Heat = m × Δ T × S H Specific heat of water, SH = 4.184 J/g 0 C 4.184 J g 0 C or g 0 C 4.184 J Conversion factor for the equality between cal and kcal: 1 kJ 1000 J or 1000 J 1kJ Heat required to melt ice, can be calculated from the heat of fusion of water, which is the heat added to convert 1. g of ice to water at 0.0 0 C. Heat of fusion of ice = 334. J/g Writing the equalities and conversion factors: 1 g H 2 O ( s → l ) = 334 J 334 J 1 g or 1 g 334 J Calculations: Calculating the amount of heat absorbed when 275 g of ice melted at 0.0 0 C: 275 g Three sig figs × 334 J Three sig fig s 1 g × 1 kJ Exact number 1000 J Exact number = 91.9 Three sig figs kJ Calculating the heat required to heat up water from 0.0 0 C to 24.0 0 C: Heat = m × Δ T × S H = 275 g Three sig figs ( 24.0 0 C Three sig figs − 0.0 0 C Two sig figs ) ( 4.184 J Four sig figs g 0 C ) × 1 kJ 1000 J = 28

Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

13th Edition

ISBN: 9780134421353

Author: Karen C. Timberlake

Publisher: PEARSON

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Chapter 3, Problem 3.101CP

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To calculate: Kilojoules of heat absorbed by 275 g ice at 0.00C to convert to water at 24.0 ⁰C.

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Chapter 3 Solutions

Chemistry: An Introduction to General, Organic, and Biological Chemistry (13th Edition)

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Kilojoules of heat absorbed by 275 g ice at 0.0 0 C to convert to water at 24.0 0 C.

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To calculate: Kilojoules of heat absorbed by 275 g ice at 0.00C to convert to water at 24.0 0C.

Chapter 3 Solutions

To calculate: Kilojoules of heat absorbed by 275 g ice at 0.00C to convert to water at 24.0 ⁰C.