A metal sphere of 0.5 kg is heated to 300 o C, than dropped into a beaker holding 4.08 kg mercury at 20 o C. The temperature of the mercury stabilizes at 99o C after a couple of minutes. What metal was the sphere made of

The metal sphere was made of iron.

Explanation:

From the question,

Heat lost by the metal sphere = heat gained by the mercury.

Cm(t₁-t₃) = C'm'(t₃-t₂) Equation 1

Where C = specific heat capacity of the metal sphere, m = mass of the metal sphere, C' = specific heat capacity of the mercury, m' = mass of the mercury.

t₁ = initial temperature of the metal sphere, t₂ = initial temperature of the mercury, t₃ = Final temperature of the mixture.

Make C the subject of the equation

C = C'm'(t₃-t₂)/m(t₁-t₃) Equation 2

Given: m = 0.5 kg, m' = 4.08 kg, t₁ = 300 °C, t₂ = 20 °C, t₃ = 99 °C

Constant: C' = 140 J/kgK.

Substitute into equation 2

C = [140×4.08×(99-20)]/[0.5×(300-99)]

C = 45124.8/100.5

C = 449.0 J/kgK.

Hence from the specific heat capacity data sheet, the metal is suspected to  be  made of iron.

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