Question

500.g of brass (c=.0920 cal/g°C) at 200.°C and 300.g of steel (c=.115 cal/g°C) at 150.°C are added to 900.g of water (c=1 cal/g°C) in a 150.g Aluminum pan (c=.220 cal/g°C), both of which are at 20.0°C. Find the common equilibrium temperature.

Answer 1

Let the equilibrium temperature be T

$\Delta Q_{total}=0$

$\therefore \sum mc\Delta T=0$

$500*0.0920(200-T)+300*0.115(150-T)+900*1(20-T)+150*0.220(20-T)=0$

$-46T+9200-345T+5775-900T+18000-33T+660=0$

1013.5T=33035

T=$32.60\degree C$