Question

at the local county fair, you watch as a blacksmith drops a 0.50kg iron horseshoe into a bucket containing 25kg of water. If the initial temperature of the horseshoe is 450C and the initial temp of water is 23C, what is the equilibrium temp of the system?

Answer 1

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 440C, and the initial temperature of the water is 22C, what is the equilibrium temperature of the system? Assume no heat is exchanged with the surroundings. >>

0.50(Ci)(440 - T) = 27(Cw)(T - 22))

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."

I trust that you can do this on you own.

For part B of the problem, the working equation will be

1(Cs)(440 - T) = 27(Cw)(T - 22)

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.

You get equilibrium temp to be 24