Answer:
0.862 J/gºC
Explanation:
The following data were obtained from the question:
Mass of metal (Mₘ) = 50 g
Initial temperature of metal (Tₘ) = 100 °C
Mass of water (Mᵥᵥ) = 400 g
Initial temperature of water (Tᵥᵥ) = 20 °C
Equilibrium temperature (Tₑ) = 22 °C
Specific heat capacity of water (Cᵥᵥ) = 4.2 J/gºC
Specific heat capacity of metal (Cₘ) =?
The specific heat capacity of the metal can be obtained as follow:
Heat lost by metal = MₘCₘ(Tₘ – Tₑ)
= 50 × Cₘ × (100 – 22)
= 50 × Cₘ × 78
= 3900 × Cₘ
Heat gained by water = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
= 400 × 4.2 × (22 – 20)
= 400 × 4.2 × 2
= 3360 J
Heat lost by metal = Heat gained by water
3900 × Cₘ = 3360
Divide both side by 3900
Cₘ = 3360 / 3900
Cₘ = 0.862 J/gºC
Therefore, the specific heat capacity of the metal is 0.862 J/gºC
Organisms that share many derived characteristics are probably more closely related
Explanation:
there is 2 nitrogen but if you mean nitrate is 6
