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Building a bronze stature we make a mold and pour in the liquid bronze when the bronze hardens we remove the mold. The mold is made of 3 Kg of steel and the statue has a mas of 1 Kg. The specific heat of bronze is 360 and the specific heat of steel is 165 If we cool the statue and the mold from 500 degrees C by dropping it into a bucket with 10 Kg of water at 20 degrees C what will the final temperature of the water be? (ignore any steam). Specific heat of water is 4186.

So far I have:

(165)(3 kg)(change in temp) + (360)(1 kg)(change in temp) = (4186)(10 kg)(change in temp)

Am I on the right track? Any tips for further help?

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1 Answer 1

Looks ok so far, but keep in mind that on the left hand side, you have change in temp of the hot metals (which are initially at the same temp), and on the right you have the change in temp of the water.

Since you have the initial temp in both cases, you can write both changes in temp in terms of the final temp, and solve the equation for that value.

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