Solution Found!
A 30.14-g stainless steel ball bearing at 117.82°C is
Chapter 6, Problem 7PE(choose chapter or problem)
A 30.14-g stainless steel ball bearing at 117.82°C is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.44°C. If the specific heat of the ball bearing is \(0.474\mathrm{\ J}/\mathrm{g}\cdot^{\circ}\mathrm{C}\), calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity.
Questions & Answers
QUESTION:
A 30.14-g stainless steel ball bearing at 117.82°C is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.44°C. If the specific heat of the ball bearing is \(0.474\mathrm{\ J}/\mathrm{g}\cdot^{\circ}\mathrm{C}\), calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity.
ANSWER:
Step 1 of 2
The goal of the problem is to calculate the final temperature of the water.
Given:
Mass of stainless steel = 30.14 g
Temperature stainless steel ball bearing = 117.82°C
Volume of water = 120.0 mL
Temperature of water = 18.44°C
Specific heat of the ball bearing =
Treating the calorimeter as an isolated system (no heat lost to the surroundings), we write
or
We know the formula to calculate heat change:
where C = heat capacity
1 mL = 1g.
Hence, mass of water = 120 g
Specific heat of water =
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