Final temperature?

Question

If 0.20 kg of iron at 100°C is added to 1.00 kg of water at 20.0°C, what is the final temperature of the system?

 

Please explain

Answer 1

Assume the system is isolated, meaning all the heat lost by the warmer body is absorbed by the cooler body.  We also need the specific heats of iron and water, which are found in tables:

 

m1 = mass of iron = 0.2 kg

 

T1 = initial temperature of the iron = 100 degrees C = 373 K (degrees Kelvin)

 

c1 = specific heat of iron = 450 J / (kg K)

 

m2 = mass of the water = 1 kg

 

T2 = initial temperature of the water = 20 degrees C = 293 K

 

c2 = specific heat of water = 4180 J / (kg K)

 

Tf = final temperature attained by both the iron and the water

 

If the system is isolated, then

 

m1*c1*(Tf - T1) + m2*c2*(Tf - T2) = 0

 

Solve the equation above for Tf:

 

Tf = (m1*c1*T1 + m2*c2*T2) / (m1*c1 + m2*c2)

 

Tf = {(0.2 kg)[450 J / (kg K)](373 K) + (1 kg)[4180 J / (kg K)](293 K)} / {(0.2 kg)[450 J / (kg K)] + (1 kg)[4180 J / (kg K)]}

 

Tf = 1258310 / 4270 K = 294.7 K = 21.7 degrees C