Asilver cube with an edge length of 2.28 cm2.28 cm and a gold cube with an edge length of 2.66 cm2.66 cm are both heated to 89.7 ∘c89.7 ∘c and placed in 100.5 ml 100.5 ml of water at 19.3 ∘c19.3 ∘c . what is the final temperature of the water when thermal equilibrium is reached?

30.1º C

Explanation:

Assuming no heat transfer except the one exchanged between the silver and gold cubes with the water, the heat lost by the metal cubes must be exactly same as the one gained by the water.

If the heat exchange process is due only to conduction, we can write the equation for this type of process, as follows:

Q = c m ∆T

where c= specific heat capacity, in J/kg.ºC

As we have already said, the heat lost by the metals, when reached to a final thermal equilibrium, must be equal to the heat gained by the water:

QAg + QAu = QH2O

Now, we need to find out the masses of the silver and gold cubes first.

We know that both are cubes, and we know the edge length, so we can get the volume, which multiplied by the density, gives us the mass.

Then, we can apply the heat transfer equation, as follows:

QAg= cAg . δAglAg3 (ti -tfn) = 230 J/kg.ºC . 10,500 kg/m3(0.0228)3 m3 (89.7ºC- tfn)

QAu= cAu . δAulAu3 (ti -tfn) = 130 J/kg.ºC . 19,300 kg/m3(0.0266)3 m3 (89.7ºC- tfn)

The sum of these two equations must be equal to the one below:

QH2O= cH2O . δH2o VH2O (ti -tfn) = 4180 J/kg.ºC . 1000 kg/m3 0.0001 m3 (tfn-19.3ºC)

Solving for tfn, we finally get:

tfn = 14,868 / 493.82 = 30.1ºC

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray