Explanation Given The volume of hot water in the cup is 500 mL . The initial temperature of water is 83 ° C . The mass of ice in the cup is 20 g . The temperature of ice cube is − 20 ° C . The specific heat capacity of ice is 2.03 J/g ⋅ C . The specific heat capacity of water is 4.184 J/g ⋅ C . The density of liquid water is 1.00 g/mL . The mass of liquid water in the cup is calculated by using the formula, Density = Mass Volume Mass = Density × Volume Substitute the values of density and volume in above equation to calculate the mass of water. Mass of H 2 O = 1.00 g/mL × 500 mL = 500 g The amount of heat released/absorbed to increase/decrease the temperature of ice and water from − 20 ° C to 0 ° C and 83 ° C to 0 ° C is determined by the formula, Q = m s ( T f − T i ) ……(1) Where, Q is the amount of heat. m is the mass of the substance. s is the specific heat capacity of the substance. T f is the final temperature. T i is the initial temperature. Substitute the values of m , s , T f and T i for hot water in equation (1). Q water = 500 g × 4.184 J/g ⋅ C × ( 0 − 83 ) ° C = − 173636 J Substitute the values of m , s , T f and T i for ice in equation (1). Q ice = 20 g × 2

Chemistry: The Central Science (14th Edition)

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Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

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Chapter 19, Problem 107IE

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To determine: The final temperature of water in the given cup.

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Chapter 19, Problem 106IE

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Chemistry: The Central Science (14th Edition)

Ch. 19.1 - The process of iron being oxidized to make iron...Ch. 19.1 - At 1 atm pressure, CO2(s) sublimes at 78oC. Is...Ch. 19.2 - Prob. 19.2.1PE Ch. 19.2 - Prob. 19.2.2PE Ch. 19.3 - Prob. 19.3.1PE Ch. 19.3 - Prob. 19.3.2PE Ch. 19.3 - Prob. 19.4.1PE Ch. 19.3 - Prob. 19.4.2PE Ch. 19.4 - Prob. 19.5.1PE Ch. 19.4 - Using the standard molar entropies in Appendix C,...

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