A 435−g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J...

Question

J/⁰C, what is the final temperature of the system (c of copper = 0.387 J/g·°C)?

Answer 1

Heat lost by copper MUST equal heat gained by water plus heat gained by the vessel

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

(435 g)(0.387 J/gº)(89.5º - Tf) = (159 g)(4.184 J/gº)(Tf - 22.8º) + (10.0 J/º)(Tf - 22.8)

15,067 - 168Tf = 665Tf - 15,168 + 10Tf - 228

843Tf = 30,463

Tf = 36.1ºC (assuming the temperature of the water and the vessel were in equilibrium at the start)