High School

A silver cube with an edge length of 2.28 cm and a gold cube with an edge length of 2.66 cm are both heated to 89.7 ∘C and placed in 100.5 mL of water at 19.3 ∘C. What is the final temperature of the water when thermal equilibrium is reached?

Answer :

The final temperature of the water when thermal equilibrium is reached is approximately 30.7 °C. This is obtained by calculating the heat gained

final temperature of the water when thermal equilibrium is reached can be determined using the principle of heat transfer.

First, we need to calculate the heat gained or lost by each object. The amount of heat gained or lost can be calculated using the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Let's calculate the heat gained or lost by the silver cube. The mass of the silver cube can be calculated using the formula m = ρV, where ρ is the density and V is the volume. The density of silver is approximately 10.5 g/cm³. The volume of the cube can be calculated as V = (edge length)3.

Using the given edge length of 2.28 cm, we can calculate the volume of the silver cube as V = (2.28 cm)³ = 11.40 cm³.

The mass of the silver cube is then m = (10.5 g/cm³)(11.40 cm³) = 119.7 g.

Next, we calculate the heat gained or lost by the silver cube using Q = mcΔT. The specific heat capacity of silver is approximately 0.24 J/g·°C.

Given that the initial temperature of the silver cube is 89.7 °C and the final temperature of the water is unknown, we can write the equation as Qsilver = (119.7 g)(0.24 J/g·°C)(Tfinal - 89.7 °C).

Now, let's calculate the heat gained or lost by the gold cube. Using the same process as above, we find that the mass of the gold cube is 164.1 g.

The specific heat capacity of gold is approximately 0.13 J/g·°C.

Given that the initial temperature of the gold cube is 89.7 °C, we can write the equation as Qgold = (164.1 g)(0.13 J/g·°C)(Tfinal - 89.7 °C).

Since heat gained by the water equals the sum of the heat lost by the silver cube and the gold cube, we have Qwater = Qsilver + Qgold.

Now we can substitute the equations and solve for the final temperature of the water.

(100.5 mL)(1 g/mL)(4.18 J/g·°C)(Tfinal - 19.3 °C) = (119.7 g)(0.24 J/g·°C)(Tfinal - 89.7 °C) + (164.1 g)(0.13 J/g·°C)(Tfinal - 89.7 °C).

Simplifying and solving the equation, we find that the final temperature of the water when thermal equilibrium is reached is approximately 30.7 °C.

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