Explanation Given: Mass of the ice present initially is m solid ice = 10.0 g . The heat capacities of ice and steam are C solid ice = 2.09 J/(g ° C) and C steam = 2.01 J/(g ° C) , respectively. Initial temperature of ice is T initial = − 10.0 ° C . Final temperature of steam is T final = 110 ° C . Explanation: The amount of heat required to heat a substance in a single state is given as follows: q = m C Δ T Here, m is mass of the substance, C is heat capacity of the substance, and Δ T is the change in temperature. Ice will get heated till it changes to liquid water at 0.0 ° C . Calculate the change in temperature for solid ice as follows: Δ T = T melting point − T initial = 0.0 ° C − ( − 10.0 ° C ) = 10.0 ° C Calculate the amount of heat required to heat the solid ice as follows: q 1 = m solid ice C solid ice Δ T = ( 10.0 g ) × ( 2.09 J/(g ° C) ) × ( 10 .0 ° C ) = 209 J × ( 1 kJ 1000 J ) = 0. 209 kJ Now, solid ice will change its phase to liquid water. The amount of heat required accompanying the phase change of ice is given as follows: q = n Δ H fus Here, n is the number of moles of ice and Δ H fus is the heat of fusion. Calculate the amount of heat required accompanying the phase change of ice by substituting the values as follows: q 2 = n Δ H fus = ( 10.0 g 18.02 g/mol ) × 6.02 kJ/mol =3 .34 kJ Now, water will get heated till it changes to steam at 100.0 ° C . Calculate the change in temperature for water as follows: Δ T = T boiling point − T = 100.0 ° C − ( 0.0 ° C ) = 100

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Chapter 11, Problem 71E

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To determine: The amount of heat required to convert 10.0 g of ice at −10.0 ° C into steam at 110.0 °C.

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Chapter 11 Solutions

Chemistry: A Molecular Approach Selected Solutions Manual, Books a la Carte Edition

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The amount of heat required to convert 10.0 g of ice at − 10.0 ° C into steam at 110.0 ° C .

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To determine: The amount of heat required to convert 10.0 g of ice at −10.0 ° C into steam at 110.0 °C.

Chapter 11 Solutions