Chapter 17: Problem 5
The background temperature of the universe is a) \(6000 \mathrm{~K}\). b) \(288 \mathrm{~K}\). c) \(3 \mathrm{~K}\). d) \(2.725 \mathrm{~K}\). e) \(0 \mathrm{~K}\).
Chapter 17: Problem 5
The background temperature of the universe is a) \(6000 \mathrm{~K}\). b) \(288 \mathrm{~K}\). c) \(3 \mathrm{~K}\). d) \(2.725 \mathrm{~K}\). e) \(0 \mathrm{~K}\).
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Get started for freeA steel antenna for television broadcasting is \(501.9 \mathrm{~m}\) tall on a summer day, when the outside temperature is \(28.09^{\circ} \mathrm{C} .\) The steel from which the antenna was constructed has a linear expansion coefficient of \(13.89 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1} .\) On a cold winter day, the temperature is \(-15.91{ }^{\circ} \mathrm{C} .\) What is the change in the height of the antenna?
The volume of \(1.00 \mathrm{~kg}\) of liquid water over the temperature range from \(0.00^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) fits reasonably well to the polynomial function \(V=1.00016-\left(4.52 \cdot 10^{-5}\right) T+\left(5.68 \cdot 10^{-6}\right) T^{2}\), where the volume is measured in cubic meters and \(T\) is the temperature in degrees Celsius. a) Use this information to calculate the volume expansion coefficient for liquid water as a function of temperature. b) Evaluate your expression at \(20.0^{\circ} \mathrm{C}\), and compare the value to that listed in Table \(17.3 .\)
An iron horseshoe at room temperature is dunked into a cylindrical tank of water (radius of \(10.0 \mathrm{~cm}\) ) and causes the water level to rise by \(0.250 \mathrm{~cm} .\) When the horseshoe is heated in the blacksmith's stove from room temperature to a temperature of \(7.00 \cdot 10^{2} \mathrm{~K},\) worked into its final shape, and then dunked back into the water, how much does the water level rise above the "no horseshoe" level (ignore any water that evaporates as the horseshoe enters the water)? Note: The linear expansion coefficient for iron is roughly that of steel: \(11.0 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)
Some textbooks use the unit \(\mathrm{K}^{-1}\) rather than \({ }^{\circ} \mathrm{C}^{-1}\) for values of the linear expansion coefficient; see Table \(17.2 .\) How will the numerical values of the coefficient differ if expressed in \(\mathrm{K}^{-1}\) ?
Two solid objects, \(A\) and \(B\), are in contact. In which case will thermal energy transfer from \(A\) to \(B\) ? a) \(\mathrm{A}\) is at \(20^{\circ} \mathrm{C},\) and \(\mathrm{B}\) is at \(27^{\circ} \mathrm{C}\) b) \(\mathrm{A}\) is at \(15^{\circ} \mathrm{C},\) and \(\mathrm{B}\) is at \(15^{\circ} \mathrm{C}\) c) \(\mathrm{A}\) is at \(0^{\circ} \mathrm{C},\) and \(\mathrm{B}\) is at \(-10^{\circ} \mathrm{C} .\)
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