Question
Calculate the increase in entropy of the Universe when you add $20.0 \mathrm{g}$ of $5.00^{\circ} \mathrm{C}$ cream to $200 \mathrm{g}$ of $60.0^{\circ} \mathrm{C}$ coffee. Assume that the specific heats of cream and coffee are both $4.20 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$
Step 1
We do this by adding 273 to the Celsius temperature. So, the temperature of the cream, $T_1$, is $5.00^{\circ}C + 273 = 278K$ and the temperature of the coffee, $T_2$, is $60.0^{\circ}C + 273 = 333K$. Show more…
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Calculate the increase in entropy of the Universe when you add 20.0 g of $5.00^{\circ} \mathrm{C}$ cream to 200 $\mathrm{g}$ of $60.0^{\circ} \mathrm{C}$ coffee. Assume the specific heats of cream and coffee are both $4.20 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C} .$ You may use the result of Problem $31 .$
It can be shown that as a mass $m$ with specificheat c changes temperature from $T_{i}$ to $T_{\mathrm{f}}$ its change in entropy is $\Delta S=m c \ln \left(T_{f} / T_{\mathrm{i}}\right)$ if the temperatures are expressed in kelvin. Suppose you put 79 $\mathrm{g}$ of milk at 278 $\mathrm{K}$ into an insulated cup containing 296 $\mathrm{g}$ of coffee at 355 $\mathrm{K}$ , and that each has the specific heat of water. The system comes to an equilibrium temperature of 339 $\mathrm{K}$ (a) What is the entropy change of the milk? (b) What is the entropy change of the coffee? (c) What is the entropy change of the universe due to adding the milk to the coffee?
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