Answer:
The specific heat capacity of the metal is approximately 0.3903 J/(g·°C)
Explanation:
The mass of the sample of the unknown metal, [tex]m_m[/tex] = 134.0 g
The temperature to which the metal is raised, [tex]t_m[/tex] = 91.0°C
The mass of water into which the mass of metal is placed, [tex]m_w[/tex] = 125 g
The temperature of the water into which the metal is placed, [tex]t_w[/tex]= 25.0°C
The final temperature of the water, [tex]t_f[/tex] = 31.0°C
The specific heat capacity of water, [tex]c_w[/tex] = 4.184 J/(g·°C)
The specific heat capacity of the metal = [tex]c_m[/tex]
Therefore, by the first laws of thermodynamics we have;
The heat transferred = Heat supplied by the metal = Heat gained by the water
The heat transferred, ΔQ, is given as follows;
ΔQ = [tex]m_w[/tex]×[tex]c_w[/tex]×([tex]t_f[/tex] - [tex]t_w[/tex]) = [tex]m_m[/tex]× [tex]c_m[/tex]×([tex]t_m[/tex] - [tex]t_f[/tex])
125 × 4.184 × (31 - 25) = 134 × [tex]c_m[/tex] × (91 - 31)
∴ [tex]c_m[/tex] = (125 × 4.184 × (31 - 25))/(134 × (91 - 31)) ≈ 0.3903 J/(g·°C)
The specific heat capacity of the metal = [tex]c_m[/tex] ≈ 0.3903 J/(g·°C)
