1. Find \( \Delta W \) and \( \Delta U \) for a \( 6.0-\mathrm{cm} \) cube of iron as it is heated from 20 - C to \( 300^{\circ} \mathrm{C} \) at atmospheric pressure. For iron, \( c=0.11 \mathrm{cal} / \mathrm{g} \cdot \mathrm{C} \) and the volume coefficient of thermal expansion is \( 3.6 \times 10^{-5}{ }^{0} \mathrm{C} \) \( { }^{-1} \). The mass of the cube is \( 2000 \mathrm{~g} \). 2. A motor supplies \( 0.4 \mathrm{hp} \) to stir \( 6 \mathrm{~kg} \) of water. Assuming that all the work goes into heating the water by friction losses, how long will it take to increase the temperature of the water \( 6^{\circ} \mathrm{C} \) ? 3. Compute the entropy change of \( 5.00 \mathrm{~g} \) of water at \( 100^{\circ} \mathrm{C} \) as it changes to steam at \( 100^{\circ} \mathrm{C} \) under standard pressure. 4. Heat in the amount of \( 100 \mathrm{~kJ} \) is transferred out of a reservoir that is sustained at \( 500 \mathrm{~K} \). Determine the resulting entropy change of the reservoir. Is the reservoir's entropy increased or decreased? [Hint. Heat-out is negative.]
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Calculate the change in entropy when $1 \mathrm{~kg}$ water at $27^{\circ} \mathrm{C}$ is converted into superheated steam at $200^{\circ} \mathrm{C}$ under constant atmospheric pressure. Specific heat capacity of liquid water is $4180 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$; temperature dependence of specific heat capacity of steam is given by the relation $(1670+0.49 T) \mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1}$ at $T$ degree kelvin. Take latent heat of steam as $23 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}$. (Ans: $\left.7521 \mathrm{~J} \mathrm{~kg}^{-1}\right)$
(a) After 6.00 $\mathrm{kg}$ of water at $85.0^{\circ} \mathrm{C}$ is mixed in a perfect thermos with 3.00 $\mathrm{kg}$ of ice at $0.0^{\circ} \mathrm{C},$ the mixture is allowed to reach equilibrium. When heat is added to or removed from a solid or liquid of mass $m$ and specific heat capacity $c$ , the change in entropy can be shown to be $\Delta S=m c \ln \left(T_{\mathrm{f}} / T_{\mathrm{i}}\right),$ where $T_{\mathrm{i}}$ and $T_{\mathrm{f}}$ are the initial and final Kelvin temperatures. Using this expression and the change in entropy for melting, find the change in entropy that occurs. $(\mathrm{b})$ Should the entropy of the universe increase or decrease as a result of the mixing process? Give your reasoning and state whether your answer in part (a) is consistent with your answer here.
Find $\Delta W$ and $\Delta U$ for a $6.0-\mathrm{cm}$ cube of iron as it is heated from $20^{\circ} \mathrm{C}$ to $300{ }^{\circ} \mathrm{C}$ at atmospheric pressure. For iron, $c=0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}$ and the volume coefficient of thermal expansion is $3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}$. The mass of the cube is $1700 \mathrm{~g}$. Given that $\Delta T=300^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=280^{\circ} \mathrm{C}$ $$ \Delta Q=c m \Delta T=\left(0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(1700 \mathrm{~g})\left(280{ }^{\circ} \mathrm{C}\right)=52 \mathrm{kcal} $$ To find that the work done by the expansion of the cube, we need to determine $\Delta V$. The volume of the cube is $V=(6.0 \mathrm{~cm})^{3}=216 \mathrm{~cm}^{3} .$ Using $(\Delta V) / V=\beta \Delta T$, $$ \Delta V=V \beta \Delta T=\left(216 \times 10^{-6} \mathrm{~m}^{3}\right)\left(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(280{ }^{\circ} \mathrm{C}\right)=2.18 \times 10^{-6} \mathrm{~m}^{3} $$ Then, assuming atmospheric pressure to be $1.0 \times 10^{5} \mathrm{~Pa}$, $$ \Delta W=P \Delta V=\left(1.0 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(2.18 \times 10^{-6} \mathrm{~m}^{3}\right)=0.22 \mathrm{~J} $$ But the First Law tells us that $$ \begin{aligned} \Delta U &=\Delta Q-\Delta W=(52000 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})-0.22 \mathrm{~J} \\ &=218000 \mathrm{~J}-0.22 \mathrm{~J} \approx 2.2 \times 10^{5} \mathrm{~J} \end{aligned} $$ Notice how very small the work of expansion against the atmosphere is in comparison to $\Delta U$ and $\Delta Q .$ Often $\Delta W$ can be neglected when dealing with liquids and solids.
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