Question

50 gram of water at 0°C is mixed with an equal mass of water at 80°C. Calculate the resultant increase in entropy

Answer 1

Given,

mass of the water = 50gm

Temperature of the water $T_1=0^\circ C$

mass of the second water =50gm

Temperature of the water $(T_2)=80^\circ C$

Let the final temperature of the water be t.

Now, applying the conservation of energy,

Heat gain = Heat loss

$50s(t-0)=50s(80-t)$

$t=40^\circ C$

Change in entropy of $0^\circ C$ water = $\frac{\Delta Q}{t}=\frac{50\times 40}{40+273}=6.39J/k$

Similarly change in entropy of $80^\circ C$ water,

$=-\frac{50\times 40}{40+273}=-6.39J/k$

Hence the total change in entropy=0J/k