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
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 25oC when its temperature reduces to 0oC,
Heat absorbed when 300g ice at -30oC increases to
0oC,
Heat required for ice to water phase transition at 0oC,
Here Q3 alone >Q1 so water temperature will drop to 0oC
The residual heat after bringing the ice at 00C is=21000-18972=2028J
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!!!
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