Given:
Metal:
Mass of the metal, m₁ = 14.9 g
Specific heat, c₁ cal/(g-°C), unknown
Water:
Mass, m₂ = 75.0 g
Specific heat, c₂ = 1.0 cal/(g-°C)
Initially:
Temperature of the metal, T₁ = 100°C
Temperature of the water, T₂ = 20°C
Finally:
Temperature of metal and water, T₃ = 28°C
Assume that there is no energy loss to the system.
Therefore
m₁c₁(T₃ - T₁) + m₂c₂(T₃ - T₂) = 0
[tex](14.9 \, g)*(c_{1} \, \frac{cal}{g-^{o}C})*(28-100 \,^{o}C) \\ + (75.0\, g)*(1.0\, \frac{cal}{g-^{o}C})*(28-20\, ^{o}C) \\ =0[/tex]
-1072.8c₁ + 600 = 0
c₁ = 0.5593 cal/(g-°C)
Answer:
The specific heat of the metal is 0.56 cal/(g °C)
