Edit to “Solution”

astrosander edited
revision #12607 newer →
@@ -13,7 +13,7 @@Solution
I. Heating the water: As $\delta Q=cmdT$:
$$
−\Delta S_{1} =\int_{T_{1}}^{T_{2}}\frac{\delta Q}{T} = cm\int_{T_{1}}^{T_{2}}\frac{dT}{T} = cm\cdot ln\frac{T_{2}}{T_{1}}
+\Delta S_{1} =\int_{T_{1}}^{T_{2}}\frac{\delta Q}{T} = cm\int_{T_{1}}^{T_{2}}\frac{dT}{T} = cm\cdot \ln\frac{T_{2}}{T_{1}}
$$
Where $T_{1} = 293~\text{K}$ and $T_{2} = 373~\text{K}$ and $c = 4.2~\mathrm{\frac{kJ}{kg\cdot \text{ K}}}$ is the specific heat of water.
@@ -29,7 +29,7 @@Solution
The resulting change in entropy:
$$
−\boxed{\Delta S =cm\cdot ln\frac{T_{2}}{T_{1}} + \frac{mL}{T_{2}}}
+\boxed{\Delta S =cm\cdot \ln\frac{T_{2}}{T_{1}} + \frac{mL}{T_{2}}}
$$
$$
unchanged lines 8