Question

A hot lump of 35.9 g of aluminum at an initial temperature of 63.3 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

m(water) = 50.0 g
T(water) = 25.0 oC
C(water) = 4.184 J/goC
m(aluminium) = 35.9 g
T(aluminium) = 63.3 oC
C(aluminium) = 0.903 J/goC
T = to be calculated


Let the final temperature be T oC
use:
heat lost by aluminium = heat gained by water
m(aluminium)*C(aluminium)*(T(aluminium)-T) = m(water)*C(water)*(T-T(water))
35.9*0.903*(63.3-T) = 50.0*4.184*(T-25.0)
32.4177*(63.3-T) = 209.2*(T-25.0)
2052.0404 - 32.4177*T = 209.2*T - 5230
T= 30.1387 oC
Answer: 30.1 oC