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Bingel [31]
3 years ago
14

A metal 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 eq

uilibrium, the temperature of the water and metal was 33.5 °c. determine the specific heat capacity of the metal.
Chemistry
1 answer:
shtirl [24]3 years ago
7 0
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)

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

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