A2.0-kg piece of steel with a temperature of 70 degree C is submerged in 1.0 kg of water at 15 degree C. At what temperature does the water/metal system reach equilibrium? The specific heat of steel is 0.107 cal/(g*K). Specific Heat of water is 4185 J/(kg*K.( The answer is 27 C)
HOW? I keep getting 70 degree C when I do m*C*delta T (of steel) = m* C* delta T (of water)
m1=2 kg,
c1= 0.107 cal/(g*K)= 448 J/kg•K,
T1=273+70=343.15 K,
m2=1kg,
c2=4185 J/kg•K,
T2=273+15=288.15 K
Q1 =m1•c1(T1-Tₒ) = m1•c1•T1- m1•c1•Tₒ
Q2 = m2•c2• ( Tₒ-T2) = m2•c2•Tₒ- m2•c2•T2
Q1=Q2
m1•c1•T1- m1•c1•Tₒ= m2•c2•Tₒ- m2•c2•T2
m1•c1•T1+ m2•c2•T2= m1•c1•Tₒ+ m2•c2•Tₒ=Tₒ( m1•c1+ m2•c2)
Tₒ = (m1•c1•T1+ m2•c2•T2)/ ( m1•c1+ m2•c2)=
=(2•448•343.15 +1•4185•288.15)/( 2•448+1•4185)=
=(307462.4+1205907.75)/(896+4185)=
=297.8 K ≈ 25 ºC