Question

11 Calculate the electric power which must be supplied to the filament of of a light bulb operating at 3000K. The total surface area of the filament is
8×10−6m2
and its emissivity is 0.92.
20.3W
33.8W
46.4W
56.7W

12 Calculate the change in internal energy of 2kg of water at
90oC
when it is changed to
3.30m3
of steam at
100oC
. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is
2.26×106J/kg
.
4.27 MJ
3.43 kJ
45.72 mJ
543.63 J

Answer 1

Answer on Question Question #58126, Physics / Mechanics | Relativity

Task Calculate the electric power which must be supplied to the filament of a light bulb operating at 3000K. The total surface area of the filament is $8 \times 10^{-6} \mathrm{~m}^2$ and its emissivity is 0.92.

Solution

According to black body theorem luminosity (or same, power), that going from surface, can be defined by formula:

$$L = \gamma A \sigma T ^ {4}$$ $$\gamma - emissivity, A - surface area, \sigma - constant, 5.67 * 10 ^ {-8}, T - surface temperature$$ $$\begin{array}{l} L = 0.92 \cdot 5.67 \cdot 10 ^ {-8} \cdot 8 \cdot 10 ^ {-6} \cdot 3000 ^ {4} = 41.7312 \cdot 10 ^ {-14} \cdot 3 ^ {4} \cdot 10 ^ {12} = 3380.2272 \cdot 10 ^ {-2} \\ \approx 33.8 \, W \\ \end{array}$$

Answer $L = 33.8 \, W$

Task Calculate the change in internal energy of $2\,\mathrm{kg}$ of water at $90\,\mathrm{oC}$ when it is changed to 3.30 m³ of steam at 100°C. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is $2.26 \times 106\,\mathrm{J/kg}$.

Solution

According to thermodynamics $\mathrm{dU} = \mathrm{PdV} - \mathrm{TdS}$, where $\mathrm{dU}$ is change in internal energy, $\mathrm{P}$ - pressure, $\mathrm{dV}$ - change in volume, $\mathrm{TdS}$ - heat, that given to system.

P is constant (according to condition) $\mathrm{dV} = V_2 - V_1$

$V_1 = 2$ liters $= 0.002\,\mathrm{m}^3$ $V_2 = 3.3\,\mathrm{m}^3$

We can assume, that $\mathrm{dV} = 3.3\,\mathrm{m}^3$

$\mathrm{TdS} = Q = \mathrm{cm}(t_2 - t_1) + qm$, where $t_2 - 100\,\mathrm{C}$ (before vaporization), $t_1$ - initial temperature $(90\,\mathrm{C})$, $q$ - latent heat of vaporization, $c$ - heat capacity $(4200\,\mathrm{J/kg/C})$, $m$ - mass of water.

$$\begin{array}{l} Q = 4200 \cdot 2 \cdot 10 + 2 \cdot 2.26 \cdot 10 ^ {6} = 4.604 \cdot 10 ^ {6} = 4.6 \cdot 10 ^ {6} \\ P d V = 100000 \cdot 3.3 = 330000 \\ d U = P d V - Q = 0.33 \cdot 10 ^ {6} - 4.6 \cdot 10 ^ {6} = - 4.27 \cdot 10 ^ {6} \\ \end{array}$$

Answer

$$d U = - 4.27 \cdot 10 ^ {6} \, J$$

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