500 g of Al is heated to 100.0°C and dropped into a l00.0 g glass beaker that contains 200.0 g of H₂O. The temperature of the glass and water is 20.0°C. What is the final temperature? (Specific heat of Al = 0.29 J/g °C, specific heat of glass = 0.84 J/g °C)

Question

Quizplus · Accepted Answer

First, we need to calculate the heat gained or lost by each substance:

Heat gained or lost by Al: q = mAl x Cal x ΔT = (0.5 kg)(0.29 J/g°C)(Tf - 100°C)

Heat gained or lost by glass: q = mglass x Cglass x ΔT = (0.1 kg)(0.84 J/g°C)(Tf - 20.0°C)

Heat gained or lost by water: q = mwater x Cwater x ΔT = (0.2 kg)(4.18 J/g°C)(Tf - 20.0°C)

We can assume that the heat lost by Al is equal to the heat gained by glass and water, so we can set up an equation:

(mAl x Cal x ΔT) = (mglass x Cglass x ΔT) + (mwater x Cwater x ΔT)

Simplifying, we get:

(0.5 kg)(0.29 J/g°C)(Tf - 100°C) = (0.1 kg)(0.84 J/g°C)(Tf - 20.0°C) + (0.2 kg)(4.18 J/g°C)(Tf - 20.0°C)

Solving for Tf, we get:

Tf = 30.9°C

Therefore, the final temperature is 30.9°C, which corresponds to answer choice D.

500 G of Al Is Heated to 100