Q=cmΔT
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
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
1 answer
1 answer