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 lost by aluminum and then calculate the heat gained by water and glass.

Heat lost by aluminum:
Q1 = mAl x cAl x ΔT
Q1 = 500 g x 0.29 J/g °C x (100-20) °C
Q1 = 11600 J

Heat gained by glass and water:
Q2 = mwater x cwater x ΔT + mglass x cglass x ΔT
Q2 = (200 g + 100 g) x 4.18 J/g °C x (Tf - 20) °C + 100 g x 0.84 J/g °C x (Tf - 20) °C
Q2 = (300 g x 4.18 J/g °C + 100 g x 0.84 J/g °C) x Tf - (300 g x 4.18 J/g °C + 100 g x 0.84 J/g °C) x 20 °C
Q2 = 4446 J/°C x Tf - 174240 J

Since the heat lost by aluminum is equal to the heat gained by glass and water:
Q1 = Q2
11600 J = 4446 J/°C x Tf - 174240 J
Tf = (11600 J + 174240 J) / (4446 J/°C)
Tf = 30.9 °C, therefore the final temperature is 30.9 °C.

500 G of Al Is Heated to 100