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An 80.0-gram sample of water at 10.0°C absorbs 1680 joules of heat energy. What is the final temperature of the water?A 4.00°CB 5.00°CC 15.0°CD 50.0°C

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Solution 1

To solve this problem, we need to use the formula for heat transfer:

Q = mcΔT

where: Q = heat energy absorbed or released (in joules) m = mass of the substance (in grams) c = specific heat capacity (for water it's 4.18 J/g°C) ΔT = change in temperature (in °C)

We are given: Q = 1680 J m = 80.0 g

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