A blacksmith drops a 0.60 kg iron horseshoe into a bucket containing 25 kg of water.

(a) If the initial temperature of the horseshoe is 600°C, and the initial temperature of the water is 40°C, what is the equilibrium temperature of the system? Assume no heat is exchanged with the surroundings.
?°C

The heat lost by the horseshoe equals the heat gained by th water. Write an equation that says this and solve it for the final temperature.

They should have told you what the horseshoe is made of, because you will need to know its specific heat.

1256

To find the equilibrium temperature of the system, we need to apply the principle of thermal equilibrium. This principle states that when two objects with different temperatures are in contact with each other, heat will flow from the object with the higher temperature to the object with the lower temperature until they reach thermal equilibrium.

In this case, the horseshoe and the water are in contact, so heat will flow between them until they reach the same temperature. We can calculate this temperature using the principle of heat transfer.

First, let's calculate the heat transferred by the horseshoe. We can use the heat transfer equation:

Q = mcΔT

where Q represents the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the horseshoe, the mass (m) is 0.60 kg, the specific heat capacity (c) of iron is approximately 450 J/kg°C, and the change in temperature (ΔT) is given by:

ΔT = final temperature - initial temperature = Tf - Ti

Since we want to find the equilibrium temperature, the final temperature of the horseshoe will be equal to the equilibrium temperature of the system. The initial temperature (Ti) of the horseshoe is 600°C.

Now let's calculate the heat transferred by the water. Similarly, we can use the heat transfer equation:

Q = mcΔT

For the water, the mass (m) is 25 kg, the specific heat capacity (c) of water is approximately 4186 J/kg°C, and the change in temperature (ΔT) is given by:

ΔT = final temperature - initial temperature = Tf - Ti

Since we want to find the equilibrium temperature, the final temperature of the water will also be equal to the equilibrium temperature of the system. The initial temperature (Ti) of the water is 40°C.

Now, according to the principle of thermal equilibrium, the heat transferred by the horseshoe is equal to the heat transferred by the water:

Qhorseshoe = Qwater

Therefore, we can set up an equation:

mcΔT (horseshoe) = mcΔT (water)

Substituting the given values:

(0.60 kg)(450 J/kg°C)(Tf - 600°C) = (25 kg)(4186 J/kg°C)(Tf - 40°C)

Now we can solve for Tf, the equilibrium temperature:

(0.60 kg)(450 J/kg°C)(Tf - 600°C) = (25 kg)(4186 J/kg°C)(Tf - 40°C)

270(Tf - 600) = 104650(Tf - 40)

Distribute and simplify:

270Tf - 162000 = 104650Tf - 4186000

Move all the Tf terms to one side and the constant terms to the other side:

-104380Tf = -4024000

Divide by -104380 on both sides:

Tf ≈ 38.50°C

Therefore, the equilibrium temperature of the system is approximately 38.50°C.