Coffee Cup Calorimeter Problem

Question

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction

A student heats 68.85 grams of lead to 97.93 °C and then drops it into a cup containing 76.93 grams of water at 23.24 °C. She measures the final temperature to be 25.77 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.52 J/°C.

Assuming that no heat is lost to the surroundings calculate the specific heat of lead.

Answer 1

heat lost by hot lead must equal heat gained by the cool water PLUS the heat gained by the calorimeter

q = heat

m = mass

C = specific heat

∆T = change in temperature Pb = 97.93 -25.77 = 72.16º

∆T for water and for calorimeter = 25.77 - 23.44 = 2.33º

heat lost by lead = q = mC∆T = (68.85 g)(C)(72.16) = 4999C = heat lost by lead

heat gained by water = (76.93 g)(4.184 J/g/deg)(2.33º) = 750 J = heat gained by water

heat gained by calorimeter = Ccal x ∆T = 1.52 J/º x 2.3º = 3.5 J = heat gained by calorimeter

4999 C = 750 J + 3.5 J

4999 C = 754 J

C = 0.151 J/g/º