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 1 year ago
Next > < Previous

Related Solutions

How much heat (in kJ) is evolved when 60.26 g of water cools and freezes from...
How much heat (in kJ) is evolved when 60.26 g of water cools and freezes from 59.63oC to 0oC? What is the change in enthalpy (in kJ) under standard conditions when 8.186 g of ethane is combusted?
1-Determine the amount of heat (in kJ) given off when 2.40 ×104g of NO2 is produced...
1-Determine the amount of heat (in kJ) given off when 2.40 ×104g of NO2 is produced according to the following equation: 2 NO(g) + O2(g) → 2NO2(g)ΔH = −114.6 kJ/mol Enter your answer in scientific notation. 2-Given the following thermochemical equations, NO(g) + O3(g) → NO2(g) + O2(g) ΔH = −198.9 kJ/mol O3(g) → 3/2O2(g) ΔH = −142.3 kJ/mol O2(g) →2O(g) ΔH = +495 kJ/mol determine the enthalpy change for the reaction 2NO2(g) → 2NO(g) + O2(g) 3-Determine the pH...
Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of...
Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J/mol⋅K and that the heat capacity of H2O(s) is 37.7 J/mol⋅K, calculate the heat of fusion of water at -10 ∘C. Please show work
Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of...
Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J/mol⋅K and that the heat capacity of H2O(s) is 37.7 J/mol⋅K, calculate the heat of fusion of water at -14 ∘C. I feel like I should be able to get this problem but something is just not clicking. Thank you for your help!
Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of...
Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J/mol⋅K and that the heat capacity of H2O(s) is 37.7 J/mol⋅K, calculate the heat of fusion of water at -13 ∘C.
How much energy is given off when 55g of steam at 155 C is cooled to...
How much energy is given off when 55g of steam at 155 C is cooled to water at 75 C. Given: C steam:2.0 J/gc C water:4.18 J/g C ice:2.03 J/gc deltaH(vap)=40.7 Kj/mol deltaH(fus): 6.02 Kj/mol
how many kj of heat are absorbed when 455g of water at 80.0C are heated to...
how many kj of heat are absorbed when 455g of water at 80.0C are heated to 100.0C and then completely evaporated at this tempertature? The Specific heat of water 4.184j/C, g and the molar heat of evaporation for water is 40.7kJ/mol.
how much heat (in kj) is needed to convert 866 g of ice at -10 c...
how much heat (in kj) is needed to convert 866 g of ice at -10 c to steam at 126 c? spH of ice and steam are 2.03J/g*c and 1.99J/g*c, respectively.
How much heat (in kJ) is needed to convert 866 g of ice at −10.0°C to...
How much heat (in kJ) is needed to convert 866 g of ice at −10.0°C to steam at 126.0°C? (The specific heats of ice, water, and steam are 2.03 J/g · °C, 4.184 J/g · °C, and 1.99 J/g · °C, respectively. The heat of fusion of water is 6.01 kJ/mol, the heat of vaporization is 40.79 kJ/mol.)
How much heat (in kJ) is evolved in converting 1.00 mole of steam at 145.0 °C...
How much heat (in kJ) is evolved in converting 1.00 mole of steam at 145.0 °C to ice at -50.0 °C? The heat capacity of ice is 2.09 J/g°C and that of steam is 2.09 J/g°C
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT