Calorimetry - Chemistry

Question

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 24.6 °C are mixed. The specific heat of water is 4.184 ^J/_g°C.  °C

Answer 1

Heat lost by warm water MUST equal heat gained by cooler water. Conservation of energy!!!

heat lost by warm water = q = mC∆T

m = mass = 50.0 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 100º - Tf where Tf is the final teperature

q = (50.0 g)(4.184 J/gº)(100 - Tf)

heat gained by cooler water = q = mC∆T

m = mass = 100 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = Tf - 24.6º

q = (100 g)(4.184 J/gº)(Tf - 24.6)

Setting these equal, we have...

(50.0 g)(4.184 J/gº)(100 - Tf) = (100 g)(4.184 J/gº)(Tf - 24.6)

20920 - 209.2Tf = 418.4Tf - 5146.2

627.6Tf = 26066.2

Tf = 41.5ºC (be sure to check the math...I did it in a rush)