The human body obtains 1078 kJ from a candy bar. If this energy were used to vaporize water at $100^{\circ} \mathrm{C}$, how much water in liters could be vaporized? (Assume that the density of water is $1.0 \mathrm{g} / \mathrm{mL} .)$
Added by Joaquin H.
Step 1
We know that 1 kJ = 1000 J. So, 1078 kJ = 1078 * 1000 = 1078000 J. Show more…
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The human body obtains 915 $\mathrm{kJ}$ of energy from a candy bar. If this energy were used to vaporize water at $100.0^{\circ} \mathrm{C}$ , how much water (in liters) could be vaporized? (Assume the density of water is 1.00 $\mathrm{g} / \mathrm{mL.}$ .
The human body obtains 910 kJ of energy from a candy bar. If this energy were used to vaporize water at 100.0 °C, how much water (in liters) could be vaporized? The enthalpy of vaporization of water at 100.0 °C is 40.7 kJ·mol⁻¹.
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(II) When a diver jumps into the ocean, water leaks into the gap region between the diver's skin and her wetsuit, forming a water layer about 0.5 $\mathrm{mm}$ thick. Assuming the total surface area of the wetsuit covering the diver is about $1.0 \mathrm{m}^{2},$ and that ocean water enters the suit at $10^{\circ} \mathrm{C}$ and is warmed by the diver to skin temperature of $35^{\circ} \mathrm{C}$ , estimate how much energy (in units of candy bars $=300 \mathrm{kcal}$ ) is required by this heating process.
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