A 20.94 g sample of an unknown metal is heated to 99.4 oC in a hot water bath. The metal sample is transferred to a calorimeter containing 100.0 mL of water at 22.0 oC. The thermal equilibrium temperature of the mixture (metal sample plus water) is 24.6 oC. How much heat (q) was LOST by the METAL?

1092 J

Explanation:

From the question,

Assuming no heat is lost to the surrounding

Heat lost by the metal = Heat gained by water.

H' = Cm(t₂-t₁) Equation 1

Where H' = Heat lost by the metal, C = specific heat capacity of water,  m = mass of water, t₂ = Final temperature of the mixture, t₁ = Initial temperature of water

But,

m = D'v Equation 2

Where D' = Density of water, v = volume of water.

Given: D' = 1000 kg/m³, v = 100 mL = 100/1000000 = 0.0001 m²

Substitute into equation 2

m = 1000(0.0001)

m = 0.1 kg.

Also given: C = 4200 J/kg.°C, t₁ = 22 °C, 24.6 °C

Substitute into equation 1

H' = 4200×0.1×(24.6-22)

H' = 420(2.6)

H' = 1092 J.

Hence the heat (q) lost by the metal = 1092 J

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