Question

During 4 hours one winter afternoon, when the outside temperature was 10° C, a house heated by electricity was kept at 19° C with the expenditure of 58 kwh (kilowatt·hours) of electric energy. (a) What was the average energy leakage in joules per second (watts) through the walls of the house to the environment (the outside air and ground)? watts (b) The rate at which energy is transferred between two systems due to a temperature difference is often proportional to their temperature difference. Assuming this to hold in this case, if the house temperature had been kept at 22° C (71.6° F), how many kwh of electricity would have been consumed? kwh

Answer 1

Answer:

a

The average energy leakage is  $A= 14500J/s$

b

The KWh of electricity consumed is  $E_1=70.43Kwh$

Explanation:

From the question we are told that

      The temperature outside $T_o = 10^o$C

       The The duration is  $t =4 hours = 4 *3600 =14400s$

       The Annual energy expenditure is  $E = 58kwh = 58*1000*3600 = 2088*10^5J$

        The temperature of the house is $T_h = 24^oC$

The average energy leakage(A) can be mathematically obtained by this formula

                            $A = \frac{E}{t}$

                               $= \frac{2088*10^5}{14400}$

                               $A= 14500J/s$    

Given that the house temperature $T_h =27^oC$

  The electrical energy consumed at $T_h_1 =24^oC$ is $E = 58Kwh$

   The electrical energy consumed at $T_h_2 = 27^0C$ is $E_1 = ?$

Since energy is proportional to $(T-T_{ambient})$

Where $T_{ambient }$ is the temperature of outside ($T_0$)

The expression of the relationship between E and $E_1$ is mathematically represented as

               $\frac{E}{E_1} = \frac{T_h_1 -T_{ambiant}}{T_h_2 - T_{ambiant}}$

Substituting values

                $\frac{E}{E_1} = \frac{24-11}{27-11}$

                      $=0.8235$

 Now making $E_1$ the subject of the equation

        $E_1 = \frac{E}{0.8235}$

             $= \frac{58}{0.8235}$

             $E_1=70.43Kwh$

             

Answer 2

a. The average energy leakage in Joules per second (Watts) through the walls of the house to the environment (the outside air and ground) is 14,500 Watts.

b. The amount of electricity that would have been consumed at 22°C (71.6°F) is 77.33 kwh.

Given the following data:

• Time = 4 hours

• Outside temperature = 10°C

• House temperature = 19°C

• Energy consumption = 58 kwh

Conversion:

Time = 2 hours to seconds = $4 \times 60 \times 60 = 14400\;seconds$

Energy = 58 kwh to Joules = $58\times 1000\times 60\times60=208.8 \times 10^6\;Joules$

a. To determine the average energy leakage in Joules per second (Watts) through the walls of the house to the environment (the outside air and ground):

During the four (4) hour interval, the quantity of heat energy that leaked out of the house is equal to the quantity of heat energy put into the house because both the initial temperature and final temperature of the house are the same.

Hence, there is no change in the thermal energy of the house.

So, the average energy leakage is given by the formula:

$P_{ave} = \frac{Energy}{Time}$

Substituting the given parameters into the formula, we have;

$P_{ave} = \frac{208.8\times 10^6}{14400}\\\\P_{ave} = 14500\;J/s$

Average energy leakage = 14,500 Watts.

b. To determine how many kwh of electricity would have been consumed:

At 19°C, energy consumption ($E_i$) = 58 kwh

At 22°C, the energy consumption is $E_f$

Note: The rate of energy flow out of the house in part (a) is directly proportional to $T_h -T_{out}$.

We would calculate the constant of proportionality as follows:

$k = \frac{T_h_i -T_{out}}{T_h_f -T_{out}} \\\\k=\frac{19-10}{22-10} \\\\k=\frac{9}{12}$

k = 0.75

For energy consumption ($E_f$):

$E_f = \frac{E_i}{k} \\\\E_f = \frac{58}{0.75} \\\\E_f = 77.33\;kwh$

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