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A 1.00-kg block of copper at 100°C is placed in an in- sulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0°C. Find the entropy change of (a) the cop- per block, (b) the water, and (c) the universe.

User David Yue
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1 Answer

3 votes

Answer:

the entropy change of the copper block = - 117.29 J/K

the entropy change of the water = 138.01 J/K

the entropy change of the universe = 20.72 J/K

Step-by-step explanation:

For Copper block:

the mass of copper block
(m_c) = 1.00 kg

Temperature of block of copper
(T_c) = 100°C

= (100+273)K

= 373K

Standard Heat capacity for copper
(C_c) = 386 J/kg.K

For water:

We know our volume of liquid water to be = 4.00 L

At 0.0°C Density of liquid water = 999.9 kg/m³

As such; we can determine the mass since :
density = (mass)/(volume)

∴ the mass of 4.00 L of liquid water at 0.0°C will be its density × volume.

= 999.9 kg/m³ ×
(4)/(1000)m^3

= 3.9996 kg

so, mass of liquid water
(m_w) = 3.9996 kg

Temperature of liquid water
(T_w) at 0.0°C = 273 K

Standard Heat Capacity of liquid water
(C_w) = 4185.5 J/kg.K

Let's determine the equilibruium temperature between the copper and the liquid water. In order to do that; we have:


m_cC_c \delta T_c =m_wC_w \delta T_w


1.00*386*(373-T_\theta)=3.996*4185.5*(T _\theta-273)


386(373-T_\theta)=16725.26(T_\theta-273)


(373-T_\theta)=(16725.26)/(386) (T_\theta-273)


(373-T_\theta)=43.33 (T_\theta-273)


(373-T_\theta)=43.33 T_\theta-11829.09


373+11829.09=43.33 T_\theta+T_\theta


12202.09 =43.33T_\theta


T_\theta= 275.26 K

∴ the equilibrium temperature = 275.26 K

NOW, to determine the Entropy change of the copper block; we have:


(\delta S)_(copper)=m_cC_cIn((T_\theta)/(T_c) )


(\delta S)_(copper)=1.0*386In((275.26)/(373) )


(\delta S)_(copper)=-117.29 J/K

The entropy change of the water can also be calculated as:


(\delta S)_(water)=m_wC_wIn((T_\theta)/(T_w) )


(\delta S)_(water)=3.9996*4185.5In((275.26)/(373) )


(\delta S)_(water)=138.01J/K

The entropy change of the universe is the combination of both the entropy change of copper and water.


(\delta S)_(universe)=(\delta S)_(copper)+(\delta S)_(water)


(\delta S)_(universe)=(-117.29+138.01)J/K


(\delta S)_(universe)=20.72J/K

User Parag Badala
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