Chapter 6: Problem 10
In the equation \(w=-P \Delta V,\) why is there a negative sign?
Chapter 6: Problem 10
In the equation \(w=-P \Delta V,\) why is there a negative sign?
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Get started for freeCombustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for $\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\( \)(-2755 \mathrm{kJ} / \mathrm{mol}),\( and \)\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\( calculate \)\Delta H$ for the reaction $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$
If the internal energy of a thermodynamic system is increased by $300 . \mathrm{J}\( while 75 \)\mathrm{J}$ of expansion work is done, how much heat was transferred and in which direction, to or from the system?
Quinone is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in certain enzymes allows plants to take water, carbon dioxide, and the energy of sunlight to create glucose. A 0.1964 -g sample of quinone $\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)$ is burned in a bomb calorimeter with a heat capacity of 1.56 \(\mathrm{kJ} / \mathrm{C}\) . The temperature of the calorimeter increases by \(3.2^{\circ} \mathrm{C}\) . Calculate the energy of combustion of quinone per gram and per mole.
Consider the dissolution of \(\mathrm{CaCl}_{2} :\) $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ} $$ An 11.0 -g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 $\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .$
The best solar panels currently available are about 19\(\%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1 \text { kilowatt }\) hour; \(1 \mathrm{kW}=1000 \mathrm{J} / \mathrm{s}\) ). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 132 for the energy rate supplied by the sun.)
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