A blacksmith cools a piece of iron of mass 1.4 kg, heated to a temperature of 430°C, by putting it into a bucket containing 15 kg water at 26 degrees * C . Find the final temperature of the water. Ignore the heat gained by the bucket itself and any steam which may be emitted. Specific heat of water and iron are 4200 and 450 Jkg^ -1 K^ -1 .
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Answers
Answer:
We assume that the water-horseshoe system is thermally isolated (insuIated) from the environment for the short time required for the horseshoe to cool off and the water to warm up. Then the total energy input from the surroundings is zero, as expressed by
Q
Fe
+Q
water
=0
(mcΔT)
F
e+(mcΔT)
water
=0
M
Fe
C
Fe
(T−600
0
C)+m
w
c
w
(T−25.0
0
C)=0
Note that the first term in this equation is a negative number of joules, representing energy lost by the originally hot subsystem, and the second term is a positive number with the same absolute value, representing energy gained by heat by the cold stuff. Solving for the final temperature gives
T=
m
Fe
c
Fe
+m
w
c
w
m
w
c
w
(25.0
0
C)+m
Fe
c
Fe
(600
0
C)
Substituting c
w
=4186J/kg.
0
C and C
Fe
=448J/kg.
0
C and suppressing units,we obtain
T=
(1.50)(448)+(20.0kg)(4186)
(20.0)(4186)(25.0
0
C)+(1.50)(448)(600
0
C)
29.6
0
C