what is the final temperature when a 3.0 kg gold bar at 99 degrees Celsius is dropped into 0.22 kg of water at 25 degrees Celsius?

TF=47

To find the final temperature when the gold bar is dropped into water, we can use the principle of heat transfer. The equation for heat transfer is:

Q = mcΔT

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

First, let's find the heat transferred from the gold bar to the water:

Qgold = mgold * cgold * ΔTgold

Given that the mass of the gold bar (mgold) is 3.0 kg and the specific heat capacity of gold (cgold) is 129 J/kg°C, and the initial temperature of the gold bar (ΔTgold) is 99°C - 25°C = 74°C, we can calculate Qgold.

Next, let's find the heat transferred to the water:

Qwater = mwater * cwater * ΔTwater

Given that the mass of water (mwater) is 0.22 kg and the specific heat capacity of water (cwater) is 4186 J/kg°C, and the initial temperature of the water (ΔTwater) is 25°C - 99°C = -74°C (negative because the water is gaining heat), we can calculate Qwater.

To find the final temperature, we can assume that the total heat lost by the gold bar is gained by the water:

Qgold = -Qwater

Using this equation, we can solve for the final temperature (Tfinal):

mgold * cgold * ΔTgold = -mwater * cwater * ΔTwater

Tfinal = Tinitial + (Qgold / (mwater * cwater))

Now, let's plug in the known values and calculate Tfinal:

Tfinal = (99°C * 3.0 kg * 129 J/kg°C) / (0.22 kg * 4186 J/kg°C) + 25°C

After performing the calculation, we get the final temperature.

call specific heat of gold G

call specific heat of water W
(I am not about to get a book out)

3 (G) (99-T) = .22 (W)(T-25)

T will be pretty close to boiling.

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