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A 15.3 gram piece of lead at 115.1 °C is placed in a calorimeter containing water at 21.9 °C. If the temperature at equilibrium is 24.5 °C, what is the mass of the water?
Report your answer in grams to two (2) sig figs.
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Question & Answer: A 15.3 gram piece of lead at 115.1 °C is placed in a calorimeter containing water at 21.9 °C…..
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Expert Answer
Answer
Heat relesed by Lead ,qr = m × ∆T × C
mass of lead,m = 15.3g
Temperature difference of lead , ∆T = 24.5℃ – 115.1℃ = – 90.6℃
Heat capacity of Lead ,C = 0.128J/g℃
qr = 15.3g ×( – 90.6℃ ) × 0.128 ( J/g ℃ )
= – 177.4J
Heat released by lead = Heat absorbed by water
Heat absorbed by water ( qa) = m × ∆T × C
mass of water ,m = ?
Temperature difference of water , ∆T = 24.5℃ – 21.9℃ = 2.6℃
Heat capacity of water, C = 4.184( J/g ℃)
177.4 J = m × 2.6℃ × 4.184( J/g ℃ )
m = 177.4J/2.6℃ × 4.184(J/g℃)
= 16.31g
so, mass of water = 16g
