# Question & Answer: A 15.3 gram piece of lead at 115.1 °C is placed in a calorimeter containing water at 21.9 °C…..

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?

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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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Heat relesed by Lead ,qr = m × ∆T × C

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