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sergey [27]
3 years ago
8

Suppose an electric toaster oven is on for 30 minutes. Roughly (to 30% accuracy) how much energy is used by the oven? (a) 5 kWh

(b) 3.6 MJ (c) 5 Wh (d) 2 million Joules
Engineering
1 answer:
katrin2010 [14]3 years ago
7 0

Answer:

Roughly, an electric toaster oven on for 30 minutes uses: d) 2 million Joules of energy

Explanation:

An average toaster oven uses around 1.2 kW (or 1200 W) of power.

If it is on for 30 minutes, we can calculate how much energy it will use, knowing that energy equals power times time (that the machine is using power), taking into account units (kWh=kW*h and J=W*s):

E_{kWh}=1.2kW*30 min*\frac{1 h}{60 min}=0.6kWh\\E_{J}=1200W*30 min*60\frac{s}{min}=2160000 J

As we assumed the power of the toaster, it is enough to say that an electric toaster oven uses roughly 2 million Joules, as there is only an 8% (\frac{|difference between values|}{value used} *100=\frac{|160000 J|}{2000000J} *100) difference between our calculated value and the rough estimation.

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The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
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Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

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