Thermochemistry

Question

Nickel pieces weighing 7.00 g total (specific heat capacity = 0.44 J/g. ⁰C) and a 14.00 g silver bar (specific heat capacity = 0.24 J/g. ⁰C) are heated to 120⁰ C. The objects are placed into 85.0 g of ethylene glycol (specific heat capacity = 2.20 J/g. ⁰C) at 18 ⁰ C. Calculate the final temperature of the system (assuming no heat lost to the surroundings).

Answer 1

heat lost by the hot metal = heat gained by the cool ethylene glycol

heat = q = mC∆T where m = mass; C= specific heat; ∆T = change in temperature

For nickel: q = (7.00 g)(0.44 J/gºC)(120 - Tf)

For silver: q = (14.00 g)(0.24 J/g/ºC)(120 - Tf)

For EG: q = (85.0 g)(2.20 J/g/ºC)(Tf - 18)

Setting heat lost by hot metals equal to heat gained by cool EG, we can write the following:

(7.00 g)(0.44 J/gºC)(120 - Tf) + (14.00 g)(0.24 J/g/ºC)(120 - Tf) = (85.0 g)(2.20 J/g/ºC)(Tf - 18)

370 - 3.08Tf + 403 - 3.36Tf = 187Tf - 3366

4139 = 193Tf

Tf = 21.45ºC