During three hours one winter afternoon, when the outside temperature was ${\mathbf{0}}^{\mathbf{0}}\mathit{C}\left({32}^{0}F\right)$, a house heated by electricity was kept at ${\mathbf{20}}^{\mathbf{0}}\mathit{C}$${68}^{0}F$with the expenditure of(kilowatt hours) of electric energy. What was the average energy leakage in joules per second through the walls of the house to the environment (the outside air and ground)?

The rate at which energy is transferred between two systems is often proportional to their temperature difference. Assuming this to hold in this case, if the house temperature had been kept at, ${\mathbf{25}}^{\mathbf{0}}\mathit{C}\mathbf{\left(}{\mathbf{77}}^{\mathbf{0}}\mathit{F}\mathbf{\right)}$how manyof electricity would have been consumed?

Internal energy is a property or state function in thermodynamics that defines the energy of a material in the absence of capillary effects and external electric, magnetic, and other fields.

The temperature difference determines the pace at which heat energy is transmitted between two systems,$P\infty \nabla T$ where the value of P is the rate of transfer of heat energy and the value $\nabla T$of is the temperature difference.

Power is the average energy leaked per second that is given by:

$P=\frac{E_1}{t}$.

Substitute $E_1=45kWh$and $t=3h$into the above formula.

$$\begin{gathered} P=\frac{45kWh}{3h} \\ =15kW \\ =\left(15kW\right)\left(\frac{1000J/s}{1kW}\right) \\ =15,000J/s \end{gathered}$$

Therefore, the power is 15,000J/s .

Find the difference in temperature of inside and outside in first case.

$$\begin{gathered} \nabla T_1=T_1-T_0 \\ ={20}^{0}C-0^{0}C \\ ={20}^{0}C \end{gathered}$$

Find the difference in temperature of inside and outside in the second case.

$$\begin{gathered} \nabla T_2=T_2-T_0 \\ ={25}^{0}C-0^{0}C \\ ={25}^0 \end{gathered}$$

Write the equation for electrical energy spent in the second situation.

$$\begin{gathered} P\propto \nabla Tn \\ \frac{E}{t}\propto \nabla Tn \\ E\propto \nabla Tn \end{gathered}$$

Divide $E_2$by $E_1$by using the obtained result.

$$\begin{gathered} \frac{E_2}{E_1}=\frac{\nabla T_2}{\nabla T_1} \\ {E}_2=\left(\frac{\nabla T_2}{\nabla T_1}\right)E_1 \end{gathered}$$

Substitute $\nabla T_2={25}^{0}C$$\nabla T_1={20}^{0}C$, and $E_1=45kWh$into the obtained equation.

role="math" localid="1657949205869" $$\begin{gathered} E_2=\left(\frac{{25}^{0}C}{{20}^{0}C}\right)\left(45kwh\right) \\ =56.3kwh \end{gathered}$$

Therefore, the energy consumed is $56.3kwh$.