Question

An irregular lump of an unknown metal has a measured density of 3.05 g/mL. The metal is heated to a temperature of 171 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.3 mL, and the temperature is recorded as 48.7 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings

Answer 1

density of metal = 3.05 g/mL

mass of metal = 25 mL x 3.05 g/mL = 76.25 g

mass of water = 34.3 g ( density of water = 1 g/mL)

Q = mcdT

m = mass

c = specific heat

specfific heat of water =  4.184 J/ (g. oC)

At thermal equilibrium,

Heat lost by metal = Heat gained by water

Hence,

mc (t1-t) = mc (t-t2)

76.25 g x c_metal x ( 171-48.7) = 34.3 g x 4.184 x (48.7 - 25)

c_metal = 0.365 J/ (g. oC)

Therefore,

specific heat of the unknown metal =  0.365 J/ (g. oC)