Question

A silver cube with an edge length of 2.36 cm and a gold cube with an edge length of 2.67 cm are both heated to 87.9 ∘C and placed in 108.0 mL of water at 19.7 ∘C . What is the final temperature of the water when thermal equilibrium is reached?

Answer 1

Hence, the final temperature is 28.3 °C .

Step-by-step explanation:

Given, the edge length of the silver cube = 2.36 cm

The volume of the silver cube =
`(Edge/ length)^3` =
`(2.36)^3` cm³ = 13.144256 cm³

Given, the edge length of the gold cube = 2.67 cm

The volume of the gold cube =
`(Edge/ length)^3` =
`(2.67)^3` cm³ = 19.034163 cm³

Density is defined as:-

`\rho=(Mass)/(Volume)`

or,

`Mass={\rho}* Volume`

So, Density of silver = 10.5 g/cm³

Thus, Mass of the silver cube = 10.5 g/cm³ * 13.144256 cm³ = 138.0147 g

So, Density of gold = 19.3 g/cm³

Thus, Mass of the gold cube = 19.3 g/cm³ * 13.144256 cm³ = 253.6841 g

So, Density of water = 1 g/cm³

Given, Volume = 108.0 mL = 108.0 cm³

Thus, Mass of the water = 1 g/cm³ * 108.0 cm³ = 108.0 g

Heat gain by water = Heat lost by gold + Heat lost by silver

Thus,

`m_(water)* C_(water)* (T_f-T_i)=-m_(gold)* C_(gold)* (T_f-T_i)-m_(silver)* C_(silver)* (T_f-T_i)`

Where, negative sign signifies heat loss

Or,

`m_(water)* C_(water)* (T_f-T_i)=m_(gold)* C_(gold)* (T_i-T_f)+m_(silver)* C_(silver)* (T_i-T_f)`

For water:

Mass = 108.0 g

Initial temperature = 19.7 °C

Specific heat of water = 4.184 J/g°C

For gold:

Mass = 253.6841 g

Initial temperature = 87.9 °C

Specific heat of water = 0.1256 J/g°C

For silver:

Mass = 138.0147 g

Initial temperature = 87.9 °C

Specific heat of water = 0.2386 J/g°C

So,

`108.0* 4.184* (T_f-19.7)=253.6841* 0.1256* (87.9-T_f)+138.0147* 0.2386* (87.9-T_f)`

`108* \:4.184\left(T_f-19.7\right)=31.86272296\left(-T_f+87.9\right)+32.93030742\left(-T_f+87.9\right)`

`451.872T_f-8901.8784=5695.30737 -64.79303038T_f`

`516.66503 T_f=14597.18577`

`T_f=(14597.18577)/(516.66503)`

`T_f = 28.25270\ ^0C`

Hence, the final temperature is 28.3 °C .