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Chapter 7: Problem 96

A \(7.26 \mathrm{kg}\) shot (as used in the sporting event, the shot put) is dropped from the top of a building \(168 \mathrm{m}\) high. What is the maximum temperature increase that could occur in the shot? Assume a specific heat of \(0.47 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\) for the shot. Why would the actual measured temperature increase likely be less than the calculated value?

Short Answer

Expert verified
The maximum temperature increase that could occur in the shot can be calculated by finding the gravitational potential energy at the top of the building and then converting that energy into a temperature increase using the shot's specific heat. The actual temperature increase is likely less than this calculated maximum due to energy losses during the fall and upon impact.

Step by step solution

01

Calculate the gravitational potential energy.

First, calculate the gravitational potential energy using the formula \(PE = m.g.h\). In this case, \(m\) is the mass of the shot, which is equal to \(7.26 \mathrm{kg}\) and \(h\) is the height from which the shot is dropped, which is \(168 \mathrm{m}\). \(g\) refers to the acceleration due to gravity, and we typically use a standard value of \(9.8 \mathrm{m/s}^2\) for this on Earth. The potential energy equation thus becomes \(PE = 7.26 \mathrm{kg} * 9.8 \mathrm{m/s}^2 * 168 \mathrm{m}\).
02

Convert this energy into a temperature increase.

Next, convert the gravitational potential energy into thermal energy using the specific heat capacity formula \(\Delta \theta = \frac {Q} {mc}\), where \(Q\) is the heat absorbed (or potential energy in this case), \(m\) is the mass and \(c\) is the specific heat. Substitute the calculated potential energy into the formula alongside with the known values of \(m\) and \(c\) to find the maximum increase in temperature. Here, \(m = 7.26 \mathrm{kg}\) or \(7260 \mathrm{g}\), \(c = 0.47 \mathrm{Jg}^{-1}\) \(^{\circ}\mathrm{C}^{-1}\), and \(Q\) is the potential energy calculated in previous step.
03

Address why actual temperature would be less.

In practice, the actual temperature increase would likely be less than this calculated value, because not all of the gravitational potential energy will be converted into thermal energy. Some energy may be lost due to air resistance while the shot falls, and more energy will be lost when it hits the ground, as sound, vibration, and potentially even light. These energy losses would reduce the amount of energy that could be converted into heat.

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Most popular questions from this chapter

A 0.75 g sample of \(\mathrm{KCl}\) is added to \(35.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) in a Styrofoam cup and stirred until it dissolves. The temperature of the solution drops from 24.8 to \(23.6^{\circ} \mathrm{C}\) (a) Is the process endothermic or exothermic? (b) What is the heat of solution of KCl expressed in kilojoules per mole of KCl?

A 0.205 g pellet of potassium hydroxide, \(\mathrm{KOH}\), is added to \(55.9 \mathrm{g}\) water in a Styrofoam coffee cup. The water temperature rises from 23.5 to \(24.4^{\circ} \mathrm{C}\). [Assume that the specific heat of dilute \(\mathrm{KOH}(aq)\) is the same as that of water.] (a) What is the approximate heat of solution of \(\mathrm{KOH}\) expressed as kilojoules per mole of \(\mathrm{KOH}?\) (b) How could the precision of this measurement be improved without modifying the apparatus?

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.

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) gas \(\left(d=1.83 \mathrm{kg} / \mathrm{m}^{3}\right)\) is used in most gas grills. What volume (in liters) of propane is needed to generate \(273.8 \mathrm{kJ}\) of heat? $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.9 \mathrm{kJ} \end{array}$$
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