Question

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.

Answer 1

Heat relesed by Lead ,q_r = 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℃

  q_{r = 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