chemistry homework

Question

A 483−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/

°

C, what is the final temperature of the system (c of copper = 0.387 J/g

·

°

C)?

Answer 1

The heat lost by the metal must equal the heat absorbed/gained by the water and the vessel containing the water. The equation for heat lost and heat gained is q = mC∆T

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

heat lost by copper = q = (483 g)(0.387 J/gº)(89.5º - Tf)

heat gained by water = q = (159 g)(4.1845 J/gº)(Tf - 22.8º)

heat gained by container = (10.0 J/º)(Tf - 22.8º)

(483 g)(0.387 J/gº)(89.5º - Tf) = (159 g)(4.1845J/gº)(Tf - 22.8º) + (10.0 J/º)(Tf - 22.8º)

16,729 -187Tf = 665Tf - 15,168 + 10Tf - 228

862Tf = 32,125

Tf = 37.3ºC = final temperature