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5 June, 22:43

If 455 J of heat is transferred to 25.0g of water at 45.0 degrees Celsius, what is the final temperature of the water?

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  1. 5 June, 23:04
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    Answer:49.3

    Explanation:4.1j/g c * 25g * (t2-45c) = 455j

    T2-45c = 455j/4.1j/g c * 25g

    455/104.6

    45+4.3 = 49.3 celsius
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