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choli [55]
3 years ago
15

Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station

must get rid of. The only way that the station can exhaust thermal energy is by radiation, which it does by using thin, 1.8-m-by-3.6-m panels that have a working temperature of about 6C. How much power is radiated from each panel? Assume that the panels are in the shade so that the absorbed radiation will be negligible. Assume that the emissivity of the panel is 1.0.
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

4462.0927 W

Explanation:

\epsilon = Emissivity of the panel = 1

\sigma = Stefan-Boltzmann constant = 5.67\times 10^{-8}\ W/m^2K^4

T = Temperature = (273.15+6)

Area of the panel is given by

A=2\times 1.8\times 3.6\\\Rightarrow A=12.96\ m^2

The power radiated is given by

P=\epsilon \sigma AT^4\\\Rightarrow P=1\times 5.67\times 10^{-8}\times 12.96\times (273.15+6)^4\\\Rightarrow P=4462.0927\ W

The power radiated from each panel is 4462.0927 W

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Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
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Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

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