Answer:
The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C
Explanation:
<u>Given data;</u>
edge length of silver, a = 2.38 cm = 0.0238 m
edge length of gold, a = 2.79 cm = 0.0279 m
final temperature of silver, t = 81.9 ° C
final temperature of Gold, t = 81.9 ° C
initial temperature of water, t = 19.6 ° C
volume of water, v = 109.5 mL = 0.0001095 m³
<u>Known data:</u>
density gold 19300 kg/m³
density silver 10490 kg/m³
density water 1000 kg/m³
specific heat gold is 129 J/kgC
specific heat silver is 240 J/kgC
specific heat water is 4200 J/kgC
<u>Calculated data</u>
Apply Pythagoras theorem to determine the side of each cube;
Silver cube;
let L be the side of the silver cube
Taking the cross section of the cube (form a right angled triangle), the edge length forms the <em>hypotenuse side</em>.
L² + L² = 0.0238²
2L² = 0.0238²
L² = 0.0238² / 2
L² = 0.00028322
L = √0.00028322
L = 0.0168
Volume of cube = L³
Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³
Gold cube;
let L be the side of the gold cube
L² + L² = 0.0279²
2L² = 0.0279²
L² = 0.0279² / 2
L² = 0.0003892
L = √0.0003892
L = 0.0197
Volume of cube = L³
Volume of the silver cube = (0.0197)³ = 7.645 x 10⁻⁶ m³
Mass of silver cube;
density = mass / volume
mass = density x volume
mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg
Mass of Gold cube
mass of gold cube = 19300 (kg/m³) x 7.645 x 10⁻⁶ (m³) = 0.148 kg
Mass of water
mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg
Let the heat gained by cold water be Q₁
Let the heat lost by silver cube = Q₂
Let the heat lost by gold cube = Q₃
Let the final temperature of water = T
Q₁ = 0.1095 kg x 4200 J/kgC x (T–19.6)
Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)
Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)
At thermal equilibrium;
Q₁ = Q₂ + Q₃
0.1095 x 4200 (T–19.6) = 0.0497 x 240 x (81.9–T) + 0.148 x 129 x (81.9–T)
459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)
459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T
459.9T + 11.928T + 19.092T = 9014.04 + 976.9032 + 1563.6348
490.92T = 11554.578
T = 11554.578 / 490.92
T = 23.54 ⁰C
Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C
ad then 
, where 
is the sea level weight, 
the sea level accel., 
the accel. above while 
the weight above.