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A 20.0-g sample of ice at −10.0°C is mixed with 100.0 g water at 80.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H 2 O( s ) and H 2 O( l ) are 2.03 and 4.18 J/g∙°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 9, Problem 118AE
Textbook Problem
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A 20.0-g sample of ice at −10.0°C is mixed with 100.0 g water at 80.0°C. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.03 and 4.18 J/g∙°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol.

Interpretation Introduction

Interpretation: The final temperature of the water and ice mixture should be to be calculated.

Concept Introduction:

The heat capacity C is defined as the ratio of heat absorbed to the temperature change. It can be given by,

                                           C = Heat absorbedTemperature change......(1)

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

For the above equation heat is:

                                                     q = S×M×ΔT......(2)

                                                       q is heat J

                                                       M is mass of sample (g)

                                                       S is specific heat capacity (J/°C·g)

                                                        ΔT  is temperature change (C)

For the process no heat loss to the surroundings means then the heat is

                                    (absorbed) =-q (released).....(3)

Explanation of Solution

Explanation

Record the given data:

                 Specific heat capacity of ice =   2.03 J/C.g

                 Specific heat capacity of water = 4.184 J/C.g

                 Mass of water is = 100g .

                 Mass of ice = 20g

                Temperature of water = 80°C

                Temperature of ice = -10.0°C

  • The given specific heat capacities, masses and temperatures are recorded as shown above.

To calculate the required heat of ice -10°C to 0°C .

The reaction is,

                              H2O(s-10°C)H2O(s0°C)

                          q1=2.30Jg.°C×20.0×10°C=406J

  • The specific heat capacity of ice, temperature change and mass of ice are plugging in to equation 2 to give heat of ice -10°C to 0°C .
  • The heat of ice -10°C to 0°C is 406kJ .

To calculate the required heat of ice 0°C to water at 0°C .

The reaction is,

                              H2O(sol0°C)H2O(lq0°C)

Molecular weight of water is 18.02g

                           q2=×20.0×1mol18.02g×6.02kJmol=6680J

  • The enthalpy of fusion and  molecular weight of water are plugging in to equation  to give heat of ice 0°C to water at 0°C .
  • The heat ice 0°C to water at 0°C . is 6680J .

To calculate the required heat of water at 80°C in to 0°C .

The reaction is,

                              H2O(lq80°C)H2O(1q0°C)

                          q3=4.2Jg.°C×100.0×80°C=33400J

  • The specific heat capacity of water, temperature change and mass of water are plugging in to equation 2 to give heat of water 80°C to 0°C .
  • The heat of water 80°C to 0°C is 33400J

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Chemistry: An Atoms First Approach
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