Question

A 2.00-kg block of aluminum at 50.0 °C is dropped into 5.00 kg of water at 20.0 °C. What is the change in entropy during the approach to equilibrium, assuming no heat is exchanged with the environment? The specific heat of aluminum is 0.22 cal/(g∙K).

Answer 1

Given:

$m_{1} = 2000gm$

$m_{2} = 2000gm$

$T_{1} = 50^{\circ}C$ = 323 K

$T_{2} = 20^{\circ}C$ = 293 K

specific heat of aluminium, $C_{p} = 0.22 cal/(g.K)$

specific heat of water, $C_{p} = 0.22 cal/(g.K)$

Formula Used:

Q = m$C_{p}\Delta T$

Solution:

Let the temperature at equilibrium be ' $T_{e}$' K

Now at equilibrium, using the given formula:

$m_{1}C_{p}\Delta T_{e} = m_{2}C\Delta T_{e}$

$2000\times 0.22\times (323 - T_{e}) = 5000\times1 \times (T_{e} - 293)$

$323 - T_{e} = \frac{5000}{440}(T_{e} - 293)$

$323 - T_{e} = 11.36(T_{e} - 293)$

Temperature at equilibrium, $T_{e}$ =  295.513 K

Now, change in entropy(in J/K) is given by:

$\Delta s = \frac{Q}{\Delta T}$

Q = m$C_{p}\Delta T_{e}$

$Q = 2000\times 0.22\times (323 - 295.513)$ = 12094.28 cal

Now,

$\Delta s = \frac{12094.28}{295.513}$ = 40.93 cal/K = 9.78 J/K

Therefore change in entropy is given by:

$\Delta s = 9.78 J/K$