Brass has density of 8.40 g/cm 3 and heat, 0.385 J/g°C. A 15.2 cm 3 piece brass at 163°C is dropped in a container with 150.0 g water at 22.4°C. Find the final temperature of the brass-water mixture?

Answer 1

Heat lost by the brass must equal heat gained by the water

q = heat = mC∆T

m = mass

C = specific heat

∆T = change in temperature

For brass:

m = 15.2 cm3 x 8.40 g/cm3 = 128 g

C = 0.385 J/gº

q = (128 g)(0.385 J/gº)(163º - Tf) where Tf is final temperature

For water:

m = 150.0 g

C = 4.184 J/gº

∆T = Tf - 22.4º

q = (150.0 g)(4.184 J/gº)(Tf - 22.4º)

Setting these two equal, we have...

(128 g)(0.385 J/gº)(163º - Tf) = (150.0 g)(4.184 J/gº)(Tf - 22.4º)

8033 - 49.28Tf = 627.6Tf - 14058

677Tf = 22091

Tf = 32.6ºC