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Chapter 7: Problem 120

James Joule published his definitive work related to the first law of thermodynamics in \(1850 .\) He stated that "the quantity of heat capable of increasing the temperature of one pound of water by \(1^{\circ} \mathrm{F}\) requires for its evolution the expenditure of a mechanical force represented by the fall of 772 lb through the space of one foot." Validate this statement by relating it to information given in this text.

Short Answer

Expert verified
By using the conversion factor of mechanical energy to heat energy, and understanding the conversion of units and the relationship between heat and work done, we validated Joule's statement which states that the heat required to raise the temperature of water by \(1^{\circ}\) F equals the mechanical energy of a 772-lb weight falling a distance of 1 foot.

Step by step solution

01

Understanding the Conversion Factor

The conversion factor to switch between mechanical energy (in joules) and heat energy (in calories) is commonly known and is approximately 4.184 J/cal. Meaning, 1 calorie of heat energy is equivalent to 4.184 joules of mechanical energy.
02

Converting Units

As per the statement, let's first convert the pound to calories. According to the specific heat of water, 1 calorie of heat will increase the temperature of 1 gram of water by \(1^{\circ} C\). In terms of Fahrenheit, it will be \(1 \times \frac{9}{5}\) calorie will increase 1 gram of water by \(1^{\circ} F\). Now, 1 pound is roughly equal to 454 grams. Therefore, to increase the temperature of 1 pound of water by \(1^{\circ} F\), it will require \(454 \times 1\frac{9}{5}\) = \(816.6\) calories.
03

Relating Heat to Work Done

Now, we need to get this into units of work. If we know that \(1\) cal is equal to approximately \(4.184\) joules, this means that \(816.6\) cal will be equivalent to \(3416.744\) joules. As per the work-energy theorem, the work done W=Fxd. Substituting value of work done, 3416.744 J = F x 1ft. The value of F is obtained as 772 pounds which is as stated by James Joule, thus validating the statement.

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Most popular questions from this chapter

One glucose molecule, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}),\) is converted to two lactic acid molecules, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s})\) during glycolysis. Given the combustion reactions of glucose and lactic acid, determine the standard enthalpy for glycolysis. $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s})+6 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2808 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) & \mathrm{COOH}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \\ 3 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(1) & \Delta H^{\circ}=-1344 \mathrm{kJ} \end{aligned}$$

The metabolism of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) yields \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) as products. Heat released in the process is converted to useful work with about \(70 \%\) efficiency. Calculate the mass of glucose metabolized by a \(58.0 \mathrm{kg}\) person in climbing a mountain with an elevation gain of \(1450 \mathrm{m}\). Assume that the work performed in the climb is about four times that required to simply lift \(58.0 \mathrm{kg}\) by \(1450 \mathrm{m} \cdot\left(\Delta H_{\mathrm{f}}^{2} \text { of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \text { is }-1273.3 \mathrm{kJ} / \mathrm{mol} .\right)\)

A plausible final temperature when \(75.0 \mathrm{mL}\) of water at \(80.0^{\circ} \mathrm{C}\) is added to \(100.0 \mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\) is (a) \(28^{\circ} \mathrm{C} ;\) (b) \(40^{\circ} \mathrm{C} ;\) (c) \(46^{\circ} \mathrm{C} ;\) (d) \(50^{\circ} \mathrm{C}\)

The heat of solution of \(\mathrm{NaOH}(\mathrm{s})\) in water is \(-41.6 \mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} .\) When \(\mathrm{NaOH}(\mathrm{s})\) is dissolved in water the solution temperature (a) increases; (b) decreases; (c) remains constant; (d) either increases or decreases, depending on how much NaOH is dissolved.

Calculate the final temperature that results when (a) a 12.6 g sample of water at \(22.9^{\circ} \mathrm{C}\) absorbs \(875 \mathrm{J}\) of heat; (b) a 1.59 kg sample of platinum at \(78.2^{\circ} \mathrm{C}\) gives off \(1.05 \mathrm{kcal}\) of heat \(\left(\mathrm{sp} \mathrm{ht} \text { of } \mathrm{Pt}=0.032 \mathrm{cal} \mathrm{g}^{-1}\right.\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\).
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