Question

A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)

Answer 1

The final temperature of the aluminum and water system, when they reach thermal equilibrium, is 48.2 °C.

Step-by-step explanation:

When the aluminum and water reach thermal equilibrium, the heat lost by the aluminum is equal to the heat gained by the water. We can use the equation: Q = mcΔT, where Q is the heat exchanged, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the aluminum:

• mass (m) = 10.00 g
• Specific heat capacity (c) = 0.90 J/g °C
• Change in temperature (ΔT) = final temperature - initial temperature = final temperature - initial temperature of water = final temperature - 25.0 °C

For the water:

• mass (m) = 50.0 g
• Specific heat capacity (c) = 4.18 J/g °C
• Change in temperature (ΔT) = final temperature - initial temperature = final temperature - 25.0 °C

Using the equation Q = mcΔT and the fact that Q is the same for both substances, we can set up the following equation:

10.00 g * 0.90 J/g °C * (final temp - 25.0 °C) = 50.0 g * 4.18 J/g °C * (final temp - 25.0 °C)

Simplifying and solving for the final temperature:

9.00 * (final temp - 25.0 °C) = 209 J/g °C * (final temp - 25.0 °C)

Final temp - 25.0 °C = (209 J/g °C)/(9.00 J/g °C) = 23.22 °C

Final temperature = 23.2 °C + 25.0 °C = 48.2 °C