Respuesta :
Answer:
Hence, the final temperature is 28.3 °C .
Explanation:
Given, the edge length of the silver cube = 2.36 cm
The volume of the silver cube = [tex](Edge/ length)^3[/tex] = [tex](2.36)^3[/tex] cm³ = 13.144256 cm³
Given, the edge length of the gold cube = 2.67 cm
The volume of the gold cube = [tex](Edge/ length)^3[/tex] = [tex](2.67)^3[/tex] cm³ = 19.034163 cm³
Density is defined as:-
[tex]\rho=\frac{Mass}{Volume}[/tex]
or, Â
[tex]Mass={\rho}\times Volume[/tex]
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, Â
[tex]m_{water}\times C_{water}\times (T_f-T_i)=-m_{gold}\times C_{gold}\times (T_f-T_i)-m_{silver}\times C_{silver}\times (T_f-T_i)[/tex]
Where, negative sign signifies heat loss
Or, Â
[tex]m_{water}\times C_{water}\times (T_f-T_i)=m_{gold}\times C_{gold}\times (T_i-T_f)+m_{silver}\times C_{silver}\times (T_i-T_f)[/tex]
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, Â
[tex]108.0\times 4.184\times (T_f-19.7)=253.6841\times 0.1256\times (87.9-T_f)+138.0147\times 0.2386\times (87.9-T_f)[/tex]
[tex]108\times \:4.184\left(T_f-19.7\right)=31.86272296\left(-T_f+87.9\right)+32.93030742\left(-T_f+87.9\right)[/tex]
[tex]451.872T_f-8901.8784=5695.30737 -64.79303038T_f[/tex]
[tex]516.66503 T_f=14597.18577[/tex]
[tex]T_f=\frac{14597.18577}{516.66503}[/tex]
[tex]T_f = 28.25270\ ^0C[/tex]
Hence, the final temperature is 28.3 °C .
