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Study B.1.5—Kelvin temperature and kinetic energy conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

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A mass of 1.0 kg of water is brought to its boiling point of $100^{\circ} \mathrm{C}$ using an electric heater of power 1.6 kW .

The molar mass of water is $18 \mathrm{~g} \mathrm{~mol}^{-1}$ . Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas.

$E_k=\frac32(1.38\times10^{-23})(373)=7.7\times10^{-21}\ \mathrm{J}$ .
$v=\sqrt{\frac{3\times1.38\times10^{-23}\times6.02\times10^{23}\times373}{0.018}}=720\ \mathrm{m\ s^{-1}}$ .

v=\sqrt{\frac{3(1.38\times10^{-23})(6.02\times10^{23})(373)}{0.018}}=720\ \mathrm{m\ s^{-1}}

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IB Physics SL B 1 5 Kelvin Temperature and Kinetic Energy Topic Practice

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