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.

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

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

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(s 0°C)

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

• 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 406 kJ .

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

The reaction is,

                              H2O(sol 0°C)→H2O(lq 0°C)

Molecular weight of water is 18.02 g

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

• 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 6680 J .

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

The reaction is,

                              H2O(lq 80°C)→H2O(1q  0°C)

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

• 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 33400 J