What will be the final temperature of 100 g of 20 degree C water when
100g of 40degree iron nails are submerged in it?  (the specific heat
of iron is 0.12 cal/g x degree C.  Here you should equate the heat
gained by the water to the heat lost by the nails.)

Hi daisy001!!


Since no heat is lost externaly (a simplicication used in problems
like this one) all the heat lost by the nails must equal the heat
gained by the water.

The heat lost or gained is:
Q = c * m * (Tf - T0)

where:
Q = change in heat energy or heat added to the sample (if negative
heat substracted);
c = specific heat;
m = mass of the sample;
(Tf - T0) = final temp less initial temp = change in temperature of sample .


For the water the heat gained is:
Q1 = c1*m1*(Tf - T0_1) = c1*m1*Tf - c1*m1*T0_1

For the iron nails the heat lost is:
Q2 = c2*m2*(Tf - T0_2) = c2*m2*Tf - c2*m2*T0_2

Since Q1 + Q2 = 0, we have that Q1 = -Q2 , then:

c1*m1*Tf - c1*m1*T0_1 = -c2*m2*Tf + c2*m2*T0_2  ==>

==> c1*m1*Tf + c2*m2*Tf = c1*m1*T0_1 + c2*m2*T0_2  ==>

==> (c1*m1 + c2*m2)*Tf =  c1*m1*T0_1 + c2*m2*T0_2  ==>

==> Tf = (c1*m1*T0_1 + c2*m2*T0_2) / (c1*m1 + c2*m2) =

       = (1*100*20 + 0.12*100*40) / (1*100 + 0.12*100) =
    
       = (2,000 + 480) / (100 + 12) =
 
       = 2480/112 =
   
       = 22.143 ºC


I hope that this helps you. Feel free to request for a clñarification
if you need it.

Regards.
livioflores-ga