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makvit [3.9K]
3 years ago
8

An engineer is investigating energy loss through windows. The windowpane of interest is 0.650 cm thick, has dimensions of 1.19 m

× 2.25 m, and has a thermal conductivity of 0.8 w/(m·°C). On a given cold day, the outside temperature is 0°C and the temperature of the interior surface of the glass is 24.00. (a) Determine the rate (in W) at which heat energy is transferred through the window (b) Determine the amount of energy (in J) transferred through the window in one day, assuming the temperature on the surfaces remains constant.
Physics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

7908.92307 W

683330953.248 J

Explanation:

k = Heat conduction coefficient = 0.8 W/(m·°C)

A = Area = 1.19\times 2.25\ m^2

l = Thickness = 0.65 cm

T_2 = 24°C

T_1 = 0°C

Rate of heat transfer is given by

Q=\frac{kA(T_2-T_1)}{l}\\\Rightarrow Q=\frac{0.8\times 1.19\times 2.25(24-0)}{0.65\times 10^{-2}}\\\Rightarrow Q=7908.92307\ W

The rate of heat transfer is 7908.92307 W

Amount of energy is given by

E=Qt\\\Rightarrow E=7908.92307\times 24\times 3600\\\Rightarrow E=683330953.248\ J

The energy transferred through the window in one day is 683330953.248 J

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Read 2 more answers
One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resist
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<h3>Answer:</h3>

(a) <u>i₁ = 0.03818 A = 38.18 mA</u>

(b) downward

(c) <u>i₂ = 0.01091 A = 10.91 mA</u>

(d) rightward

(e) <u>i₃ = 0.02727 A = 27.27 mA</u>

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(g) <u>Eₐ = 3.818 Volts</u>

<h3>Question:</h3>

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

<h3></h3><h3>Explanation:</h3>

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0  

E₂ - i₃R₃ - i₁R₁ = 0  

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0     ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0    ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

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12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

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using this value in eqn of  i₂:

i₂ = [6 - 100(0.02727)]/300

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<u>i₂ = 0.01091 A</u>

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

<u>i₁ = 0.03818 A</u>

<u></u>

(a)

<u>i₁ = 0.03818 A = 38.18 mA</u>

(b)

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Therefore, its direction will <u>downward</u>

(c)

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(d)

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Therefore, its direction will <u>rightward</u>

(e)

<u>i₃ = 0.02727 A = 27.27 mA</u>

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>leftward</u>

<u>(g)</u>

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

<u>Eₐ = 3.818 Volts</u>

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