Question

200 gr of water in a thermally insulated container. 200 gr of water is initially at 25 o C in a thermally insulated calorimeter.

a) If 50 gr of ice at –15 o C is dropped into this calorimeter what is the final temperature after thermal equilibrium is established.

b) If Instead 300 gr of ice at –30 o C is added how much ice will remain when equilibrium is reached?

c) In part (a) what is the change in entropy after the equilibrium is reached?

Answer 1

ANSWER

Part a)

From theory we know that,

E = mcT and E=mL

are used to find values, where E=energy, m=mass,T=temperature, c=specific heat capacity and L=specific latent heat.

c of water= 4200

c of Ice= 2108

Then,

Eneygy released by water = Energy absorbed by Ice to heat up

0.200*4200*(25 - T) = 0.050*2108* (T - 15)
21000 - 840T = 105.4T - 1581
19419 = 945.4T
T = 20.54 ºC = 293.69 K

Part b)

Heat released from 0.2 kg of water at 25^oC when its temperature reduces to 0^oC,

$Q_{1}=0.2\times 4200\times 25 =21000 J$
Heat absorbed when 300g ice at -30^oC increases to 0^oC,

$Q_{2}=0.3\times 2108\times 30=18972J$

Heat required for ice to water phase transition at 0^oC,

$Q_{3}=0.3\times 334000=100200J$

Here Q₃ alone >Q₁ so water temperature will drop to 0^oC

The residual heat after bringing the ice at 0⁰C is=21000-18972=2028J

$m\times 334000=2028 J$

m=6.07 g

Amount of ice = 300 - 6.07 = 293.93 g

Part c)

Entropy change is given by
dS = dQ/T => ΔS = ∫T1→T2 dQ/T
for the melting process
ΔS = ΔHf/Tm
for heating/cooling from T1 to T2 at constant heat capacity:
ΔS = m·C·ln(T2/T1)
In contradiction to (a) where you can calculate in °C because only temperature differences occur, you have ti calculate in K here, because you've temperature ratios.

So for portion of ice
ΔS_ice = m_ice·( ΔHf/Tm + C·ln(Tf/Tm) )
and for the hot water
ΔS_water = m_water·C·ln(Tf/Tb) )

The sum of the two values is the total change of entropy:
ΔS = ΔS_ice + ΔS_water
= 50g · ( 333.55J/g / 258.15K + 4.18J/gK ·ln(293.69K / 273.15K) ) + 200g · 4.18J/gK · ln(293.69K / 298.15K) )
ΔS = 67.16J/K

Regards!!!