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

Answer: a) 4.028J/s

b) 77.3kW

Explanation:

Detailed explanation and calculation is shown in the image below.