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24 July, 17:51

A large punch bowl holds 3.35 kg of lemonade (which is essentially water) at 20.0° C.

A 1.62-kg ice cube at. - 10.2°. C is placed in the lemonade.

What is the final temperature of the system, and the amount of ice (if any) remaining?

Ignore any heat exchange with the bowl or the surroundings.

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  1. 24 July, 20:16
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    Using heat balance, mass times heat capacity times temperature difference, the balance looks like:

    3.35 * 1000 g * 4.18 J / g C * (20-T) = 1.62 * 1000 g * 80 * 4.1858 J / g + 1.62 * 1000 g * 4.18/2 J / g C * (10.2 C) + 1.62 * 1000 g * 4.18 J / gC * (T-0)

    Final temperature, T is equal to 14.29 C
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