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A chunk of aluminum weighing 19.7 grams and originally at 98.01 °C is dropped into an insulated cup containing 81.3 grams of water at 21.05 °C. Assuming that all of the heat is transferred to the water, the final temperature of the water is

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Answer:

T(final) = 35°C

Explanation:

Use ∑(heat flows) = ∑q = 0

q(Al) + q(H₂O) = (mcΔT)Al + (mcΔT)H₂O = 0

[(19.7g)(0.91cal/g°C)(T(f) - 98.01°C)] + [(81.3g)(1.00cal/g°C)(T(f) - 21.05°C)] = 0

Collecting terms and solving for T(f) = 35°C final mix temperature

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