Question

A piece of unknown metal with mass 14.9 g is heated to 1000c and dropped into 75.0 g of water at 200c. the final temperature of the system is 28 degree celsius. what is the specific heat of the metal?

Answer 1

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

`(14.9 \, g)*(c_(1) \, (cal)/(g-^(o)C))*(28-100 \,^(o)C) \\ + (75.0\, g)*(1.0\, (cal)/(g-^(o)C))*(28-20\, ^(o)C) \\ =0`
-1072.8c₁ + 600 = 0
c₁ = 0.5593 cal/(g-°C)

Answer:
The specific heat of the metal is 0.56 cal/(g °C)