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, = 134.0 g
The temperature to which the metal is raised, = 91.0°C
The mass of water into which the mass of metal is placed, = 125 g
The temperature of the water into which the metal is placed, = 25.0°C
The final temperature of the water, = 31.0°C
The specific heat capacity of water, = 4.184 J/(g·°C)
The specific heat capacity of the metal =
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 = ××( - ) = × ×( - )
125 × 4.184 × (31 - 25) = 134 × × (91 - 31)
∴ = (125 × 4.184 × (31 - 25))/(134 × (91 - 31)) ≈ 0.3903 J/(g·°C)
The specific heat capacity of the metal = ≈ 0.3903 J/(g·°C)