What is the quantity of energy required to heat 50.0 grams of water from 25°C to 99°C?

To calculate the quantity of energy required to heat the water, we can use the formula for heat energy:

Q = mcΔT

Where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)

In this case:
m = 50.0 g
c = 4.18 J/g°C
The initial temperature (Tinitial) = 25°C
The final temperature (Tfinal) = 99°C
The change in temperature (ΔT) = Tfinal – Tinitial = 99°C – 25°C = 74°C

Now, we can substitute the values into the formula:

Q = (50.0 g) * (4.18 J/g°C) * (74°C)

Calculating this gives:

Q = 50.0 * 4.18 * 74 = 15460 J

To convert joules to kilojoules, we divide by 1000:

Energy in kJ = 15460 J / 1000 = 15.46 kJ

Therefore, the quantity of energy required to increase the temperature of a 50.0 gram sample of water from 25°C to 99°C is 15.46 kJ.

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