Question

256 g of aluminum at 212.5 oC are placed in an insulated container with 22.2 g of mercury at 13.8 oC. What is the final temperature for the two metals in the container? (The specific heat of mercury is 0.139 J/g oC.

Answer 1

m(mercury) = 22.2 g

T(mercury) = 13.8 oC

C(mercury) = 0.139 J/goC

m(aluminium) = 256.0 g

T(aluminium) = 212.5 oC

C(aluminium) = 0.902 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by aluminium = heat gained by mercury

m(aluminium)*C(aluminium)*(T(aluminium)-T) = m(mercury)*C(mercury)*(T-T(mercury))

256.0*0.902*(212.5-T) = 22.2*0.139*(T-13.8)

230.912*(212.5-T) = 3.0858*(T-13.8)

49068.8 - 230.912*T = 3.0858*T - 42.584

T= 210. oC

Answer: 210. oC