Final answer:
The amount of energy to increase the temperature of 20 g of water from 28.3 °C to 303.0 °C is approximately 22,832.96 J.
Step-by-step explanation:
To determine the amount of energy required to increase the temperature of 20 g of water from 28.3 °C to 303.0 °C, we can use the formula:
q = mcΔT
Where:
q = energy absorbed (in joules)
m = mass of water (in grams)
c = specific heat of water (4.184 J/g °C)
ΔT = change in temperature (in °C)
Plugging in the values:
q = (20 g) * (4.184 J/g °C) * (303.0 °C - 28.3 °C)
Simplifying the equation:
q = 20 * 4.184 * 274.7 J
Therefore, the amount of energy that must be absorbed by 20 g of water to increase its temperature from 28.3 °C to 303.0 °C is approximately 22,832.96 J.