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

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 c = 4.18 J/g°C

We need to find ΔT, which is the difference between the final temperature and the initial temperature. We can rearrange the formula to solve for ΔT:

ΔT = Q / (mc)

Substituting the given values:

ΔT = 1680 J / (80.0 g * 4.18 J/g°C) = 5°C

The change in temperature is 5°C. Since the initial temperature was 10.0°C, the final temperature is:

10.0°C + 5°C = 15.0°C

So, the correct answer is C, 15.0°C.