When 108 g of water at a temperature of 22.5?

1 answer:

3 0

<span>Î”T for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC. Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used q=mCÎ”T 11466.58 Joules = 65.1g x 4.18 J / g C x Î”T 11466.58/(65.1gx4.18)=Î”T Î”T=42.14oC So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC Î”T of second sample).</span>

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Gala2k [10]

Explanation:

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