At the local county fair, you watch as a blacksmith drops a 0.50-kg iron horsesh
ID: 1864011 • Letter: A
Question
At the local county fair, you watch as a blacksmith drops a 0.50-kg iron horseshoe into a bucket containing 27 kg of water. If the initial temperature of the horseshoe is 460 ?C, and the initial temperature of the water is 24 ?C, what is the equilibrium temperature of the system? Assume no heat is exchanged with the surroundings.Suppose the 0.50-kg iron horseshoe had been a 1.0-kg lead horseshoe instead. Would the equilibrium temperature in this case be greater than, less than, or the same as in part (a)? At the local county fair, you watch as a blacksmith drops a 0.50-kg iron horseshoe into a bucket containing 27 kg of water. If the initial temperature of the horseshoe is 460 ?C, and the initial temperature of the water is 24 ?C, what is the equilibrium temperature of the system? Assume no heat is exchanged with the surroundings.
Suppose the 0.50-kg iron horseshoe had been a 1.0-kg lead horseshoe instead. Would the equilibrium temperature in this case be greater than, less than, or the same as in part (a)?
Suppose the 0.50-kg iron horseshoe had been a 1.0-kg lead horseshoe instead. Would the equilibrium temperature in this case be greater than, less than, or the same as in part (a)?
Explanation / Answer
Use the law of conservation of energy in solving this problem, i.e.,
Heat given up by horseshoe = Heat absorbed by water
<< If the initial temperature of the horseshoe is 460C, and the initial temperature of the water is 24C, what is the equilibrium temperature of the system? Assume no heat is exchanged with the surroundings. >>
0.50(Ci)(460 - T) = 27(Cw)(T - 24))
where
Ci = specific heat of iron
T = equilibrium temperature
Cw = specific heat of water
At this point, simply obtain the specific heats of iron and water and substitute them in the above equation. After substituting these values, you can solve for the equilibrium temperature "T."
0.5*448(460-T) = 27*4186*(T-24)
=> 103040 – 224T = 113022T - 2712528
=> T = 24.86 degree Celsius
For part B of the problem, the working equation will be
1(Cs)(460 - T) = 27(Cw)(T - 24)
where
Cs = specific heat of steel
and all the other terms have been previously defined.
As in the previous problem, substitute the appropriate values of the specific heats and solve for the equilibrium temperature "T."
As soon as this new equilibrium temperature is known, you can then compare if it is greater than, less than or the same as in part A.
1*502*(460 – T) = 27*4186*(T – 24)
=> 230920 – 502T = 113022T - 2712528
=> T = 25.93 degree Celsius
Its greater.
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