Question

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.

Answer 1

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

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 : $density = \frac{mass}{volume}$

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

= 999.9 kg/m³ × $\frac{4}{1000}m^3$

= 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:

$m_cC_c \delta T_c =m_wC_w \delta T_w$

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

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

$(373-T_\theta)=\frac{16725.26}{386} (T_\theta-273)$

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

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

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

$12202.09 =43.33T_\theta$

$T_\theta= 275.26 K$

∴ the equilibrium temperature = 275.26 K

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

$(\delta S)_{copper}=m_cC_cIn(\frac{T_\theta}{T_c} )$

$(\delta S)_{copper}=1.0*386In(\frac{275.26}{373} )$

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

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

$(\delta S)_{water}=m_wC_wIn(\frac{T_\theta}{T_w} )$

$(\delta S)_{water}=3.9996*4185.5In(\frac{275.26}{373} )$

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

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

$(\delta S)_{universe}=(\delta S)_{copper}+(\delta S)_{water}$

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

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