A 11.6 g piece of metal is heated to 98°C and dropped into a calorimeter containing 50.0 g of water (specific heat capacity of water is 4.18 J/g°C) initially at 20.5°C. The empty calorimeter has a heat capacity of 125 J/K. The final temperature of the water is 28.2°C. Ignoring significant figures, calculate the specific heat of the metal.

Answer : The specific heat of the metal is,

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

where,

q = heat released by the reaction

= heat absorbed by the calorimeter

= heat absorbed by the water

c = specific heat of metal = ?

= specific heat of calorimeter =

= specific heat of water =

= mass of water = 50.0 g

m = mass of metal = 11.6 g

= final temperature =

= temperature of metal =

= temperature of water =

Now put all the given values in the above formula, we get:

Thus, the specific heat of the metal is,

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.