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You submerge 100 grams of steel in 200 grams of water. If the steel has an initial temperature of 90°C and the water has an initial temperature of 15°C, what is the final temperature of the system? The specific heat capacity of the steel is 0.15 cal/g•°C.

Sagot :

Ultimatesadsimpidn

Answer:

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

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