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Q. 3.10

Expert-verifiedFound in: Page 97

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

An ice cube (mass $30\mathrm{g}$) $0\xb0\mathrm{C}$ is left sitting on the kitchen table, where it gradually melts. The temperature in the kitchen is $25\xb0\mathrm{C}$.

(a) Calculate the change in the entropy of the ice cube as it melts into water at $0\xb0\mathrm{C}$. (Don't worry about the fact that the volume changes somewhat.)

(b) Calculate the change in the entropy of the water (from the melted ice) as its temperature rises from $0\xb0\mathrm{C}$ to $25\xb0\mathrm{C}$.

(c) Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water.

(d) Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative, or zero? Is this what you would expect?

(a) The required net change in entropy is $\Delta {S}_{1}=36.70{\mathrm{JK}}^{-1}$.

(b) Change in entropy of water is ${S}_{{\mathrm{H}}_{2}\mathrm{O}}=47.70{\mathrm{JK}}^{-1}$.

(c) Change in entropy of kitchen is $\Delta {S}_{\text{room}}=44.14{\mathrm{JK}}^{-1}$.

(d) Change in entropy of universe is $\Delta {S}_{\text{universe}}=3.56{\mathrm{JK}}^{-1}$.

Mass of ice cube, $m=30\mathrm{g}$

Initial temperaturelocalid="1647139485052" $=T=0\xb0\mathrm{C}=273\mathrm{K}$

Temperature in the kitchenlocalid="1647158003634" $=25\xb0C=298\mathrm{K}$

The change in entropy is given as:

$\Delta S=\frac{Q}{T}..........\left(1\right)$

Where,

$Q$ = heat added

role="math" localid="1647139806158" $T$ = Temperature in Kelvin

The amount of heat required to melt ice is given as:

$Q=m\times L$

Where,

$m$ = mass

$L$ = Latent heat of water = $334J{g}^{-1}$

By substituting the values in the above equation, we get,

role="math" localid="1647140235964" $Q=30\times 334\phantom{\rule{0ex}{0ex}}Q=10020J$

Now, by substituting the value of $Q$ in the equation (1), we get,

$\Delta {S}_{1}=\frac{10020}{273}\phantom{\rule{0ex}{0ex}}\Delta {S}_{1}=36.70{\mathrm{JK}}^{-1}$

Hence, the required net change in entropy is calculated as $36.70J{K}^{-1}$.

Mass of ice cube, $m=30\mathrm{g}$

Initial temperature$=T=0\xb0\mathrm{C}=273\mathrm{K}$

Temperature in the kitchenlocalid="1647158013609" $=25\xb0\mathrm{C}=298\mathrm{K}$

If an amount of heat $Q$ flows from a substance at a constant temperature, the change in entropy is given as:

$\Delta {S}_{2}={C}_{V}{\int}_{{T}_{i}}^{{T}_{f}}\frac{1}{T}dT$

Where,

${C}_{v}$ = Specific heat

$T$ = Temperature in Kelvin

The water absorbs heat as it warms up, but the temperature changes. As a result, it is preferable to employ the entropy-heat-capacity relationship, as well as the fact that the specific heat of water is roughly constant over the liquid range at $c=4.18{\mathrm{Jg}}^{-1}{\mathrm{K}}^{-1}$.

Hence, by further solving the above equation, we get:

$\Delta {S}_{2}={C}_{V}[\mathrm{ln}T{]}_{{T}_{i}}^{{T}_{f}}\phantom{\rule{0ex}{0ex}}\Delta {S}_{2}={C}_{V}\left[\mathrm{ln}\frac{{T}_{f}}{{T}_{i}}\right]$

But, we know that,

${C}_{V}=mc$

By substituting it, we get,

$\Delta {S}_{2}=mc\left[\mathrm{ln}\frac{{T}_{f}}{{T}_{i}}\right]$

Now substitute the values in the above equation:

$\Delta {S}_{2}=30\times 4.18\times \left[\mathrm{ln}\frac{298}{273}\right]\phantom{\rule{0ex}{0ex}}\Delta {S}_{2}=11J{K}^{-1}$

Thus, the total entropy increase of the water becomes:

${S}_{{H}_{2}O}=\Delta {S}_{1}+\Delta {S}_{2}\phantom{\rule{0ex}{0ex}}{S}_{{H}_{2}O}=36.70+11\phantom{\rule{0ex}{0ex}}{S}_{{H}_{2}O}=47.70J{K}^{-1}$

Hence, the change in entropy of water is calculated as $47.70J{K}^{-1}$.

Mass of ice cube, $m=30g$

Initial temperature $=T=0\xb0\mathrm{C}=273\mathrm{K}$

Temperature in the kitchen $=25\xb0\mathrm{C}=298\mathrm{K}$

If an amount of heat $Q$ flows into a substance at a constant temperature $T$, the change in entropy is given as:

localid="1647197183247" role="math" $\Delta {S}_{room}=\frac{Q}{T}$

Where,

$Q$ = heat added

$T$ = Temperature in Kelvin

Total amount of heat transferred to the water from the room's air is given as:

$Q={Q}_{ice}+{Q}_{\text{warm}}$

Where,

${Q}_{ice}$ = Heat added from ice

${Q}_{warm}$ = Heat added due to temperature change

We know that,

Total amount of heat transferred to the water from the room's air is calculated as:

$Q={Q}_{ice}+mc\Delta T$

In this case,

As calculated before in part (a), ${Q}_{ice}=10020$ and $c=4.18J{K}^{-1}{g}^{-1}$

Now, by substituting these values and the given values in the above equation, we get,

$Q=\left(10020\right)+[4.18\times 30\times (298-273\left)\right]\phantom{\rule{0ex}{0ex}}Q=13155J$

Now, by substituting this value, change in entropy can be calculated as:

$\Delta {S}_{\text{room}}=-\frac{Q}{T}\phantom{\rule{0ex}{0ex}}\Delta {S}_{\text{room}}=-\frac{13155}{298}\phantom{\rule{0ex}{0ex}}\Delta {S}_{\text{room}}=-44.14J{K}^{-1}$

Hence, the entropy change of the kitchen is calculated as $-44.14J{K}^{-1}$.

Mass of ice cube, $m=30g$

Initial temperature $=T=0\xb0\mathrm{C}=273\mathrm{K}$

Temperature in the kitchen $=25\xb0\mathrm{C}=298\mathrm{K}$

Net change of entropy of water role="math" localid="1647197282233" ${S}_{{\mathrm{H}}_{2}\mathrm{O}}=47.70{\mathrm{JK}}^{-1}$

If an amount of heat $Q$ flows into a substance at a constant temperature $T$, the change in entropy is given as:

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

Where,

$Q$ = heat added

$T$ = Temperature in Kelvin

The net change of entropy of the universe for this case can be given as:

$\Delta S=\Delta {S}_{\text{room}}+{S}_{{H}_{2}O}$

Where,

${S}_{room}$ = net change of entropy of the kitchen

${S}_{{H}_{2}O}$ = net change of entropy of water

By substituting the calculated values in the above equation, we get,

$\Delta S=-44.14+47.70\phantom{\rule{0ex}{0ex}}\Delta S=3.56J{K}^{-1}$

Hence, the net change of entropy of the universe is calculated as $3.56J{K}^{-1}$.

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