   # 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. ### Chemistry: An Atoms First Approach

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

#### Solutions

Chapter
Section ### 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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