Calorimetric equation

A 200g aluminum cylinder at 180°C was dropped into 3kg water at 25°C. At what temperature did the resulting temperature stabilize? The specific capacity of water is 4200 J/kg °C, aluminum 896 J/kg °C,

Correct answer:

t =  27.17 °C

Step-by-step explanation:

m1=3 kg t1=25 °C c1=4200 J/kg/°C  m2=200 g kg=200:1000  kg=0.2 kg t2=180 °C c2=896 J/kg/°C  Q1=Q2  m1 c1 (tt1)=m2 c2 (t2t) 3 4200 (t25)=0.2 896 (180t)  12779.2t=347256  t=12779.2347256=27.17353199  t=27.173532=27.17°C



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