Given:
[tex]m_{1} = 2000gm[/tex]
[tex]m_{2} = 2000gm[/tex]
[tex]T_{1} = 50^{\circ}C[/tex] = 323 K
[tex]T_{2} = 20^{\circ}C[/tex] = 293 K
specific heat of aluminium, [tex]C_{p} = 0.22 cal/(g.K)[/tex]
specific heat of water, [tex]C_{p} = 0.22 cal/(g.K)[/tex]
Formula Used:
Q = m[tex]C_{p}\Delta T[/tex]
Solution:
Let the temperature at equilibrium be ' [tex]T_{e}[/tex]' K
Now at equilibrium, using the given formula:
[tex]m_{1}C_{p}\Delta T_{e} = m_{2}C\Delta T_{e}[/tex]
[tex]2000\times 0.22\times (323 - T_{e}) = 5000\times1 \times (T_{e} - 293)[/tex]
[tex] 323 - T_{e} = \frac{5000}{440}(T_{e} - 293)[/tex]
[tex] 323 - T_{e} = 11.36(T_{e} - 293)[/tex]
Temperature at equilibrium, [tex]T_{e}[/tex] = 295.513 K
Now, change in entropy(in J/K) is given by:
[tex]\Delta s = \frac{Q}{\Delta T}[/tex]
Q = m[tex]C_{p}\Delta T_{e}[/tex]
[tex]Q = 2000\times 0.22\times (323 - 295.513)[/tex] = 12094.28 cal
Now,
[tex]\Delta s = \frac{12094.28}{295.513}[/tex] = 40.93 cal/K = 9.78 J/K
Therefore change in entropy is given by:
[tex]\Delta s = 9.78 J/K[/tex]
