Question

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)

Answer 1

Mass of Iron is 24.45 kg

Step-by-step explanation:

Given that:

Mass of Silver,
`m_(S)` = 8.00 kg

Mass of Gold,
`m_(G)` = 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,
`c_(G)` = 0.150 kJ/(kg K)

Specific heat of Silver-liquid,
`c_(S)` = 0.280 kJ/(kg K)

Known: Specific heat capacity of Iron,
`c_(I)` = 0.461 kJ/(kg K)

Therefore;

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

`m_(I)` x
`c_(I)` x Δθ = (
`m_(S)` x
`c_(S)` +
`m_(G)` x
`c_(G)`) x Δθ

`m_(I)` x 0.461 x (1737 - 1337) = (8 x 0.280 + 15 x 0.150) x (1337 - 333)

`m_(I)` x 0.461 x 400 = (2.24 + 2.25) x 1004

`m_(I)` x 184.4 = 4507.96

`m_(I)` =
`(4507.96)/(184.4)`

`m_(I)` = 24.4466

`m_(I)` ≅ 24.45

The mass of Iron is 24.45 kg.