A 92.9 g piece of a silvery gray metal is heated to 178.0∘C, and then quickly transferred into 75.0 mL of water initially at 24.0∘C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7∘C. Determine the specific heat and the identity of the metal. You are given that water=4.184 J GC, and the density of water is 1.00 g mL.

Q: A 92.9 g piece of a silvery gray metal is heated to 178.0∘C, and then quickly transferred into 75.0 mL of water initially at 24.0∘C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7∘C. Determine the specific heat and the identity of the metal. You are given that water=4.184 J/g.°C, and the density of water is 1.00 g/mL.

Answer:

0.1297 J/g.°C

The metal is lead.

Explanation:

Assuming no heat is lost to the surrounding,

Heat lost by the gray metal = heat gained by water

c'm' (t₁-t₃) = cm (t₃-t₁) ... Equation 1

Where, c' = specific heat capacity of the gray metal, m' = mass of the gray metal, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of the gray metal, t₂ = initial temperature of water, t₃ = temperature of the mixture,

Making c' the subject of the equation,

c' = cm (t₃-t₁) / m' (t₁-t₃) ... Equation 2

Given: m' = 92.9 g, c = 4.184 J/g.°C, t₁ = 178°C, t₂ = 24°C, t₃ = 29.7°C

But m = density of water * volume

m = 1.00*75 = 75 g

Substitute into equation 2,

c' = 4.18*75 (29.7-24) / [92.9 (178-29.7) ]

c' = 1786.95/13777.07

c' = 0.1297 J/g.°C

Hence The specific heat capacity of the metal = 0.1297 J/g.°C

The specific heat capacity of the metal is approximately equal to that of lead which is 0.13 J/g.°C. Hence the metal is lead.