A cube at 333 K contains two metals: 8.00 kg of solid Silver and 15.0 kg of solid Gold. It is placed in contact with a block of solid Iron at 1737 K. The system reaches equilibrium at 1337 K (all the silver and all the gold has melted) Find the mass of the iron. (cgold-liquid= 0.150 kJ/(kg K), csilver-liquid= 0.280 kJ/(kg K)

Question

contestada

Respuesta :

olayemiolakunle65 olayemiolakunle65 · Answer 1 · 2020-09-16T03:59:24+02:00

Answer:

Mass of Iron is 24.45 kg

Explanation:

Given that:

Mass of Silver, [tex]m_{S}[/tex] = 8.00 kg

Mass of Gold, [tex]m_{G}[/tex] = 15.0 kg

Initial temperature of Silver and Gold = 333 K

Initial temperature of Iron = 1737 K

Final temperature = 1337 K

Specific heat capacity of Gold-liquid, [tex]c_{G}[/tex] = 0.150 kJ/(kg K)

Specific heat of Silver-liquid, [tex]c_{S}[/tex] = 0.280 kJ/(kg K)

Known: Specific heat capacity of Iron, [tex]c_{I}[/tex] = 0.461 kJ/(kg K)

Therefore;

Heat lost by Iron = Heat gained by Silver + Heat gained by Gold

[tex]m_{I}[/tex] x [tex]c_{I}[/tex] x Δθ = ([tex]m_{S}[/tex] x [tex]c_{S}[/tex] + [tex]m_{G}[/tex] x [tex]c_{G}[/tex]) x Δθ

[tex]m_{I}[/tex] x 0.461 x (1737 - 1337) = (8 x 0.280 + 15 x 0.150) x (1337 - 333)

[tex]m_{I}[/tex] x 0.461 x 400 = (2.24 + 2.25) x 1004

[tex]m_{I}[/tex] x 184.4 = 4507.96

[tex]m_{I}[/tex] = [tex]\frac{4507.96}{184.4}[/tex]

[tex]m_{I}[/tex] = 24.4466

[tex]m_{I}[/tex] ≅ 24.45

The mass of Iron is 24.45 kg.