Question

A student obtained an unknown metal sample that weighed 65.3 g and at a temperature of 99.8oC, he placed it in a calorimeter containing 43.7 g of water at 25.7oC. At equilibrium the temperature of the water and metal was 34.5oC. Knowing the specific heat of the water to be 4.18 J/goC, what is the specific heat of the metal

Answer 1

0.377 J/gºC

Step-by-step explanation:

From the question given above, the following data were obtained:

Mass of metal (Mₘ) = 65.3 g

Initial temperature of metal (Tₘ) = 99.8 °C

Mass of water (Mᵥᵥ) = 43.7 g

Initial temperature of water (Tᵥᵥ) = 25.7 °C

Equilibrium temperature (Tₑ) = 34.5 °C

Specific heat capacity of water (Cᵥᵥ) = 4.18 J/gºC

Specific heat capacity of metal (Cₘ) =?

The specific heat capacity of metal can be obtained as illustrated below:

Heat lost by metal = heat gained by water.

MₘCₘ(Tₘ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

65.3 × Cₘ (99.8 – 34.5) = 43.7 × 4.18 (34.5 – 25.7)

65.3Cₘ × 65.3 = 182.666 × 8.8

4264.09Cₘ = 1607.4608

Divide both side by 4264.09

Cₘ = 1607.4608 / 4264.09

Cₘ = 0.377 J/gºC

Therefore the specific heat capacity of the metal is 0.377 J/gºC