Question

Estimate the change in the entropy of the universe due to heat escaping from your home on a cold winter day.

Short answer

The entropy change on a cold winter day can be estimated to be$8.0\times {10}^4\mathrm{J}/\mathrm{K}$.

Step-by-step solution

Given

It is given to estimate the change in the entropy of the universe due to heat escaping from your home on a cold winter day.

Hence,

Let's assume:

Power consumed from an average house on a winter day $=P=4kW=4\times {10}^{3}J/s$

Temperature inside $=T_{in}=293K$

Temperature outside$=T_{out}=275K$

Calculation

Total heat loss in a day can be calculated as:

$Q=Pt$

Where,

$P$= Power

$t$= time

Hence,

$$\begin{gathered} Q=4\times {10}^3\times 24\times 60\times 60 \\ Q=3.46\times {10}^{8}J \end{gathered}$$

Now,

Entropy gained by outdoors can be given as:

$\Delta S_{\text{out}}=\frac{Q}{T_{\text{out}}}$

By substituting the values, we get,

$$\begin{gathered} \Delta S_{\text{out}}=\frac{3.46\times {10}^8}{275} \\ \Delta S_{\text{out}}=1.26\times {10}^6\mathrm{J}/\mathrm{K} \end{gathered}$$

And,

Entropy gained by indoors can be given as:

$\Delta S_{in}=-\frac{Q}{T_{in}}$

By substituting the values, we get,

$$\begin{gathered} \Delta S_{\text{in}}=-\frac{3.46\times {10}^8}{293} \\ \Delta S_{in}=-1.18\times {10}^6\mathrm{J}/\mathrm{K} \end{gathered}$$

We know that the net entropy change is given as:

$\Delta S_{net}=\Delta S_{out}+\Delta S_{\text{in}}$

By substituting the calculated values in the above equation, we get,

$$\begin{gathered} \Delta S_{\text{net}}=\left(1.26\times {10}^6\right)-\left(1.18\times {10}^6\right) \\ \Delta S_{\text{net}}=8.0\times {10}^{4}J/K \end{gathered}$$

Final answer

Hence, the required entropy change can be calculated as$8.0\times {10}^{4}J/K$.