Posted by **AL** on Sunday, July 1, 2012 at 10:06pm.

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)

- Physics -
**Elena**, Monday, July 2, 2012 at 7:39am
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

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