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ella [17]
3 years ago
11

An electric water heater consumes 2.5 kW for 1.9 h per day.

Physics
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

The cost of running the electric water heater for one year is 55.2391 $

Explanation:

The simple rule of 3 helps to quickly solve proportionality problems when you have three known values ​​and one unknown. If two quantities are directly proportional (that is, when multiplying or dividing one of them by a number, the other is multiplied or divided respectively by the same number) the rule of three can be applied as follows:

a ⇒ b

c ⇒ x

So: x=\frac{c*b}{a}

where a, b and c are the known values ​​and x is the value you want to find out.

In this case, you can first apply the following rule of three: if 2.5 kW are consumed in 1.9 hours, in 1 hour how many kW are consumed?

kWh=\frac{1 h*2.5 kW}{1.9 h}

kWh=1.316

So an electric water heater consumes 1.316 kWh in one day. You apply another simple rule of three: if the heater in 1 day consumes 1.316 kWh, in 365 days (1 year) how many kWh are consumed?

kWh=\frac{365 days*1.316 kWh}{1 day}

kWh= 480.34

So an electric water heater consumes 480.34 kWh in a year.

If 1 kWh costs 11.5 cents, 480.34 kWh how many cents does it cost?

cost=\frac{480.34 kWh*11.5 cents}{1 kWh}

cost= 5,523.91 cents

Finally, if 100 cents is equal to 1 dollar, 5,523.91 cents, how many dollars are equal?

cost=\frac{5,5523.91 cents*1 dollar}{100 cents}

cost= 55.2391 $

<u><em> The cost of running the electric water heater for one year is 55.2391 $</em></u>

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Answer:

final-temperature = T_{f} = 252.51K

Explanation:

we can solve this problem by using the first law of thermodynamics.

    \Delta U= Q-W

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<u> C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}    (1)</u>

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