Ametal sample weighing 43.5 g and at a temperature of 100.0 °c was placed in 39.9 g of water in a calorimeter at 25.1 °c. at equilibrium, the temperature of the water and metal was 33.5 °c. determine the specific heat capacity of the metal.

The initial state of the system is comprised of
(a) A metal sample
    m₁ = 43.5 g, mass
    T₁ = 100°C, temperature
    c₁ (unknown) specific heat, J/(g-C)

(b)  Water
   m₂ = 39.9 g, mass
   T₂ = 25.1°C, temperature
   c₂ = 4.184 J/(g-C), specific heat

The final state of the system is
 M = m₁ + m₂, total mass
 T = 33.5°C, equilibrium temperature

Work in SI units. Note that changes in °C are equal to changes in °K.
Equate change in total thermal energy to zero because the energy is conserved.
m₁c₁(T-T₁) + m₂c₂(T-T₂) = 0
(43.5)*(c₁)(33.5 - 100) + (39.9)*(4.184)*(33.5 - 25.1) = 0
-2892.8c₁ + 1402.3 = 0
c₁ = 1402.3/2892.8 
    = 0.4848 J/(g-C)

The specific heat capacity of the metal is 0.485 J/(g-°C)

Atomicity occurs to prevent updates of the database occurring only partially, which can cause greater problems than rejecting the whole series outright.some are monatomic, diatomic, triatomic etc because it shows the number of atoms. for example, monatomic means it contains only one atom.

What? i don’t understand