Question

A hot lump of 39.9 g of aluminum at an initial temperature of 85.4 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

To find the final temperature of the aluminum and water, use the principle of heat transfer and the equation q = mcΔT. Set the heat for the aluminum equal to the heat for the water and solve for T_f. The final temperature is approximately 27.8°C.

Step-by-step explanation:

To find the final temperature of the aluminum and water, we can use the principle of heat transfer: q_aluminum = -q_water. The heat lost by the aluminum is equal to the heat gained by the water, assuming no heat is lost to the surroundings.

Using the equation q = mcΔT, where q is heat, m is mass, c is specific heat, and ΔT is the change in temperature, we can calculate the heat for both substances. For the aluminum, q_aluminum = (39.9 g)(0.903 J/(g·°C))(T_f - 85.4 °C). For the water, q_water = (50.0 mL)(1 g/mL)(4.18 J/(g·°C))(T_f - 25.0 °C).

Setting these two equations equal to each other and solving for T_f (the final temperature of both substances), we can find that T_f ≈ 27.8°C.

Answer 2

To find the final temperature when aluminum and water reach thermal equilibrium, we can use the equation: Mass of Aluminum x Specific Heat of Aluminum x Change in Temperature of Aluminum = Mass of Water x Specific Heat of Water x Change in Temperature of Water. Substitute the same variable for both temperatures and solve the equation to find the final temperature 'T'.

When two objects reach thermal equilibrium, their final temperature is the same. To calculate the final temperature, we can use the equation:

Mass of Aluminum x Specific Heat of Aluminum x Change in Temperature of Aluminum = Mass of Water x Specific Heat of Water x Change in Temperature of Water

Let's plug in the given values:

• Mass of Aluminum = 39.9 g
• Specific Heat of Aluminum = 0.903 J/(g·°C)
• Change in Temperature of Aluminum = Final Temperature - Initial Temperature of Aluminum = Final Temperature - 85.4 °C
• Mass of Water = 50.0 mL = 50.0 g
• Specific Heat of Water = 4.18 J/(g·°C)
• Change in Temperature of Water = Final Temperature - Initial Temperature of Water = Final Temperature - 25.0 °C

Since we're trying to find the final temperature of both the aluminum and water, we can substitute the same variable for both temperatures. Let's call it 'T'. The equation becomes:

39.9 g x 0.903 J/(g·°C) x (T - 85.4 °C) = 50.0 g x 4.18 J/(g·°C) x (T - 25.0 °C)

Solving this equation will give us the final temperature 'T'.