Question

3 Calculate the work done against external atmospheric pressure when 1 g of water changes to
1672cm3
of steam. Take the atmospheric pressure as
1.013×105Nm−2
169.3 J
342.4 J
226.2 J
143.5 J

4 An electric kettle contains 1.5 kg of water at
100oC
and powered by a 2.0 kW electric element. If the thermostat of the kettle fails to operate, approximately how long will it take for the kettle boil dry? (Take the specific latent heat of vaporization of water as
2000kJkg−1)
500s
1000s
1500s
3000s

Answer 1

Answer on Question #58114, Physics / Molecular Physics | Thermodynamics |

3 Calculate the work done against external atmospheric pressure when 1 g of water changes to 1672 cm³ of steam. Take the atmospheric pressure as 1.013×10⁵ Nm⁻²

169.3 J

342.4 J

226.2 J

143.5 J

Solution:

$$W = P (V _ {s t e a m} - V _ {w a t e r})$$ $$W = P m \left(\frac {1}{\rho_ {s t e a m}} - \frac {1}{\rho_ {w a t e r}}\right)$$ $$W = 1.013 \cdot 10^{5} \cdot 10^{-3} \left(\frac {1672 \cdot 10^{-6}}{10^{-3}} - \frac {1}{10^{3}}\right) = 169.3 \mathrm{~J}$$

Answer: 169.3 J

4 An electric kettle contains $1.5\,\mathrm{kg}$ of water at $100{}^{\circ}\mathrm{C}$ and powered by a $2.0\,\mathrm{kW}$ electric element. If the thermostat of the kettle fails to operate, approximately how long will it take for the kettle boil dry? (Take the specific latent heat of vaporization of water as $2000\,\mathrm{kJ\cdot kg^{-1}}$)

```

500s

1000s

1500s

3000s

```

**Solution:**

Needed amount of heat

$$Q = m\lambda$$

where $m$ is mass of water and $\lambda$ is specific heat of vaporization.

Then the time will be heat divided by power:

$$t = \frac{Q}{P} = \frac{1.5 \cdot 2000 \cdot 10^3}{2 \cdot 10^3} = 1500\,\mathrm{s}$$

**Answer:** 1500s

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