Question

In: Chemistry

How much heat (in kJ) is given off when 41.0g of water at 50.0°C is changed...

How much heat (in kJ) is given off when 41.0g of water at 50.0°C is changed to 41.0g of ice at -5.00°C?

The specific heat of water is 4.184 J/g·°C
The specific heat of ice is 2.087 J/g·°C
The heat of fusion of water is 6.02 kJ/mol
1 mol of water is 18.0g

Solutions

Expert Solution

Water                        

Mass    41        g         

T          50        degC  

sp heat             4.184   J/g deg

Heat of fusion 6.02     kJ/mol

heat of fusion in J/mol= 6020            

Molar mass of water   18 g/mol        

Heat of fusion in J /g 108360            J/g      

Ice                              

T is -5 deg C

sp heat =2.087            J/g deg

                                   

Calculation of heat                             

q=C x m x delta T                              

q          Heat in J                     

C         specific heat               

m         Mass in g                    

Delta T change in T               

                                   

There are following transitions in temperature                                  

                                   

                                   

                                                                      Delta T           

1st (q1)            from 50 deg C to 0 deg C       -50       deg C

2nd (q2)           From 0 deg C to 0 Deg C       0        deg C

3rd (q3)           From 0 deg C to -5.0 deg C    -5       deg C

                                   

Lets calculate q value for all these transitions                                   

q = q1+q2+q3                        

q1= (41.0 x 4.184 x -50 ) J                            

-8577.2 J                   

For q2 we use delta H in J/g                          

q2 = Delta H x mass of water                        

q2=(108360 x 41.0) J                         

=4442760 J               

Heat of freezing = - heat of fusion                            

Therefore q2 has minus sign                          

q2= -4442760             

q3 calculation is based on ice and therefore use the specific heat of ice.                             

q3=(41.0 x 2.087 x -5 ) J                               

=-427.835        J                      

q = -8577.2 - 4442760 -427.835

= -4451765.0 J

So the heat given by 41.0 g water in order to decrease its T from 50 deg C to -5 deg C would be

4451765.0 J

           

queen_honey_blossom answered 2 years ago
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