Question

1.64 mol of nickel at 150.13ºC is placed in 1.00 L of water at 25.09ºC. The final temperature of the nickel-water mixture is 26.34ºC. What is the specific heat of nickel?

a. 0.439 J K-1 mol-1
b. 0.554 J K-1 mol-1
c. 2.28 J K-1 mol-1
d. 25.7 J K-1 mol-1
e. 31.6 J K-1 mol-1

Answer 1

Heat gained or released by a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1) 

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. From this, we can calculate the what is being asked in the problem. We do as follows:

Heat gained = Heat lost
m C (Tf-Ti) = - m C (Tf-Ti) 
Since what is aksed is the specific heat capacity in units of J/mol-K, we change m into n which is the number of moles of a substance.
1.64 C (150.13 - 26.34) = (1000 /18.02) (4.18 x 18.02 )(26.34 - 25.09)
C = 25.7 J / mol-K <----------OPTION D