You submerge 100 grams of steel in 200 grams of water If the

You submerge 100 grams of steel in 200 grams of water. If the steel has an initial temperature of 80°C and the water has an initial temperature of 10°C, what is the final temperature of the system? The specific heat capacity of the steel is 0.15 cal/g•°C.

Solution

sol:

We know that heat required

Q=cmT
Q1(steel)=Q2(water)
c1•m1• (t1-t) = c2•m2• (t-t2)
628•0.1•(80-t) = 4180•0.2• (t-10)
5024-62.8t = 836t -8360
5024+8360=(836+62.8)t
t=14.9°C

Final temperature of the system is 14.9 °C