How much energy is required to boil 55.1 grams of water? heat of vaporization of water is \( 2260 \mathrm{~J} / \mathrm{g} \)

To find the total energy required to boil 55.1 grams of water, you can simply multiply the mass of the water by the heat of vaporization: \[ Energy = mass \times heat~of~vaporization = 55.1~g \times 2260~\frac{J}{g} = 124,546~J \] So, it would take approximately 124,546 Joules of energy to boil 55.1 grams of water. That's enough energy to power a small appliance for a short period! In terms of practical cooking, boiling water may seem straightforward, but remember that this energy is just the beginning! If you're making pasta or simmering soup, you'll need to keep the burner going, as some of that heat dissipates into the air or is absorbed by your pot. So think of it as keeping a cozy warm bubble pot after boiling – the fun of cooking is all in the details!