posted by Anonymous .
a 200g aluminum container has 500g of water with an initial temperature of 22 degree celsius. The following heated pieces of metal are dropped into water at the same time as follows: 300g piece of aluminum heated at 100 degree celsius; a 50g piece of copper heated at 80 degree celsius and a 40g piece of steel heated at 120 degree celsius. What will be the final temperature of the mixture?
Al: m1= 0,2 kg, t1= 22 ºC, c1 = 897 J/kg•ºC
Water: m2= 0.5 kg, t1= 22 ºC, c2= 4180 J/kg•ºC
Al: m3= 0.3 kg, t3= 100 ºC, c1=897 J/kg•ºC
Cu: m4=0.05 kg, t4=80 ºC, c4=385 J/kg•ºC
Steel: m5 =0.04 kg, t5=120 ºC, c5=466 J/kg•ºC
m1•c1•(t-t1) + m2•c2• (t-t1) =m3•c1• (t3-t)+m4•c4• (t4-t)+m5•c5v(t5-t).
Solve for „t“
Use the specific heat of metal formula for each metal to compute for the Tmix. Then, sum all the Tmix to get the Total Tmix.