15.5 g of an unknown metal at 165.0°C is dropped into 150.0mL of H2O at 23.0°C in a coffee cup calorimeter. The metal and H2O reached thermal equilibrium at 30.0°C. Calculate the specific heat capacity of the unknown metal. Specific heat of water is 4.184 J/g°C.

Specific heat capacity of metal is 2.09 j/g.°C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of metal = 15.5 g

Initial temperature = 165.0°C

Initial temperature of water = 23.0°C

Final temperature = 30.0°C

Specific heat capacity of metal = ?

Specific heat capacity of water = 4.184 J/g°C

Volume of water = 150.0 mL or 150.0 g

Solution:

Formula:

- Qm  =  +Qw

Now we will put the values in formula.

-15.5 g × c × [ 30.0°C - 165.0°C] = 150 g × 4.184 J/g°C × [ 30.0°C - 23.0°C]

15.5 g × c × 135°C = 4393.2 j

2092.5 g.°C  × c = 4393.2 j

c = 4393.2 j/2092.5 g.°C  

c = 2.09 j/g.°C  

it should be all of them because it asks if you included all of those in your answer

explanation:

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."