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+2
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Ein elektrischer Wassererhitzer leistet 1800 Watt und erhitzt Wasser in 15 min von 15°C auf 85°C.

1. Berechne die Stromkosten für die Erwärmung von 1l Wasser, wenn 1kWh 6ct kostet:

a) allgemein,

b) für die angegebenen Werte.

2. Mit welchem Wirkungsgrad arbeitet die Anlage?

18.08.2021

#1
+936
+1
18.08.2021
#2
+14776
0

War ein Irrtum. ich habe mich vertippt.

!

asinus  18.08.2021
#3
+936
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OK, passiert mal

Mathefreaker2021  19.08.2021
#4
+2524
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Greetings Asinus,

Question three (3) is missing. I included it below.

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General definitions:

$$\text {The specific heat capacity of water is } \mathrm {4.1868 J \cdot g^{-1} \cdot C°^{-1}} \hspace{1em} | \hspace{1em } \small \text {Joule per gram per degree Celsius}\\$$

1(a): Costs in general

$$\text {1800watts = 1.8kW}\\ \text {1.8kW * 6ct/kWh = 10.8ct/hour of operation }\\ \text {Operating this system for 15 minutes will cost 10.8/4 = 2.7ct }\\$$

1(b): Indicated values

$$\text {Quantity of water: 1 L = 1 Kg }\\ \text {Temperature change: Delta °C = 70°C | (85-15) }\\ \text {Power input to system: 1800 Watts | Watts/hour = 6480000J/hour }\\ \text {System Time for temperature rise: 15 minutes. }\\ \text {System operates for 15 minutes: cost of operation } \fbox {2.7ct} \hspace{1em}|\hspace{1em}\small \text {Same as costs in general }\\$$

2. Efficiency of the system:

$$\text {Input Energy required for 70} \hspace{-.1em} \mathrm {°C} \text { temperature rise for 1Kg water: }\\ E_J = 70\hspace{-.1em} \mathrm {°C} * \mathrm{1000g * 4.1868 J \cdot g^{-1} \cdot ° \hspace{-.1em} \mathrm {C^{-1}}} = \mathrm {293076J }\\ \text {@ 100% efficiency: }\\ kWh = \mathrm {\frac{293076J} {3600000J/kWh} = 0.08141 \hspace{.15em} \small kWh} \hspace{1em} | \small \text { Required energy to raise temperature by 70°C for 1 liter of water } \\ \text {Time to transfer energy to water @ 1800 Watts: } \frac {293076J}{1800J/s } = \mathrm{162.82s} = \text{2 minutes 43 seconds.} \\ \text {Cost of energy: 0.08141kWh * 6ct/kWh \aprox 0.49ct }\\ \text {Actual time to heat water: 15 minutes. }\\ \text {Energy used by system to heat water: } \frac{1.8kWh}{4} = 0.45kWh\\ \text {Efficiency of system: } \frac {0.08141kWh}{0.45kWh} \approx \fbox{18.09%}\\$$

3. The general cost of warming the cold tits of a 60kg-witch to those of a hot babe.

$$\text {Assume: 39°C temperature rise, standard human physiology, non-frozen tits, }\\ \text{ and core body temperature approximates temperature.of tits }\\ \text{Specific heat of human body: } \mathrm {3.470 \hspace{.1em}J \cdot g^{-1} \cdot ° \hspace{-.1em} C^{-1} }\\ \text {Input Energy required for 39°C temperature rise for 60kg human witch: }\\ \mathrm {E_J = 39°C * 60000g * 3.470 \hspace {.1em} J \cdot g^{-1} \cdot ° \hspace{-.1em} C^{-1} = 8119800 \hspace{.1em} J}\\ \mathrm{kWh = \frac{8119800J} {3600000J/kWh} \approx 2.26 kWh }\\ \text{Time to transfer energy @1800 Watts: } \mathrm{\frac {8119800J} {1800} = 4511s}\\ \approx 1.25 \text{ hours @100% efficiency and 6.9 hours @ 18.09% efficiency. }\\ \text{Cost: 75ct at 18.09% efficiency}\\$$

Notes: results may vary depending on the type of witch, but the tits will be much warmer.

GA

22.11.2021
bearbeitet von Gast  22.11.2021