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University Physics Volume 2

Challenge Problems

University Physics Volume 2Challenge Problems

Challenge Problems


A pendulum is made of a rod of length L and negligible mass, but capable of thermal expansion, and a weight of negligible size. (a) Show that when the temperature increases by dT, the period of the pendulum increases by a fraction αdT/2αdT/2. (b) A clock controlled by a brass pendulum keeps time correctly at 10°C10°C. If the room temperature is 30°C30°C, does the clock run faster or slower? What is its error in seconds per day?


At temperatures of a few hundred kelvins the specific heat capacity of copper approximately follows the empirical formula c=α+βT+δT−2,c=α+βT+δT−2, where α=349J/kg·K,α=349J/kg·K, β=0.107J/kg·K2,β=0.107J/kg·K2, and δ=4.58×105J·kg·K.δ=4.58×105J·kg·K. How much heat is needed to raise the temperature of a 2.00-kg piece of copper from 20°C20°C to 250°C250°C?


In a calorimeter of negligible heat capacity, 200 g of steam at 150°C150°C and 100 g of ice at −40°C−40°C are mixed. The pressure is maintained at 1 atm. What is the final temperature, and how much steam, ice, and water are present?


An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white (e=0)(e=0) except for a 5cm×8cm5cm×8cm patch in the form of the astronaut’s national flag. The patch has emissivity 0.300. The spacesuit under the patch is 0.500 cm thick, with a thermal conductivity k=0.0600W/m°Ck=0.0600W/m°C, and its inner surface is at a temperature of 20.0°C20.0°C. What is the temperature of the patch, and what is the rate of heat loss through it? Assume the patch is so thin that its outer surface is at the same temperature as the outer surface of the spacesuit under it. Also assume the temperature of outer space is 0 K. You will get an equation that is very hard to solve in closed form, so you can solve it numerically with a graphing calculator, with software, or even by trial and error with a calculator.


Find the growth of an ice layer as a function of time in a Dewar flask as seen in Exercise 1.120. Call the thickness of the ice layer L. (a) Derive an equation for dL/dt in terms of L , the temperature T above the ice, and the properties of ice (which you can leave in symbolic form instead of substituting the numbers). (b) Solve this differential equation assuming that at t=0t=0, you have L=0.L=0. If you have studied differential equations, you will know a technique for solving equations of this type: manipulate the equation to get dL/dt multiplied by a (very simple) function of L on one side, and integrate both sides with respect to time. Alternatively, you may be able to use your knowledge of the derivatives of various functions to guess the solution, which has a simple dependence on t. (c) Will the water eventually freeze to the bottom of the flask?


As the very first rudiment of climatology, estimate the temperature of Earth. Assume it is a perfect sphere and its temperature is uniform. Ignore the greenhouse effect. Thermal radiation from the Sun has an intensity (the “solar constant” S) of about 1370W/m21370W/m2 at the radius of Earth’s orbit. (a) Assuming the Sun’s rays are parallel, what area must S be multiplied by to get the total radiation intercepted by Earth? It will be easiest to answer in terms of Earth’s radius, R. (b) Assume that Earth reflects about 30% of the solar energy it intercepts. In other words, Earth has an albedo with a value of A=0.3A=0.3. In terms of S, A, and R, what is the rate at which Earth absorbs energy from the Sun? (c) Find the temperature at which Earth radiates energy at the same rate. Assume that at the infrared wavelengths where it radiates, the emissivity e is 1. Does your result show that the greenhouse effect is important? (d) How does your answer depend on the the area of Earth?


Let’s stop ignoring the greenhouse effect and incorporate it into the previous problem in a very rough way. Assume the atmosphere is a single layer, a spherical shell around Earth, with an emissivity e=0.77e=0.77 (chosen simply to give the right answer) at infrared wavelengths emitted by Earth and by the atmosphere. However, the atmosphere is transparent to the Sun’s radiation (that is, assume the radiation is at visible wavelengths with no infrared), so the Sun’s radiation reaches the surface. The greenhouse effect comes from the difference between the atmosphere’s transmission of visible light and its rather strong absorption of infrared. Note that the atmosphere’s radius is not significantly different from Earth’s, but since the atmosphere is a layer above Earth, it emits radiation both upward and downward, so it has twice Earth’s area. There are three radiative energy transfers in this problem: solar radiation absorbed by Earth’s surface; infrared radiation from the surface, which is absorbed by the atmosphere according to its emissivity; and infrared radiation from the atmosphere, half of which is absorbed by Earth and half of which goes out into space. Apply the method of the previous problem to get an equation for Earth’s surface and one for the atmosphere, and solve them for the two unknown temperatures, surface and atmosphere.

  1. In terms of Earth’s radius, the constant σσ, and the unknown temperature TsTs of the surface, what is the power of the infrared radiation from the surface?
  2. What is the power of Earth’s radiation absorbed by the atmosphere?
  3. In terms of the unknown temperature TeTe of the atmosphere, what is the power radiated from the atmosphere?
  4. Write an equation that says the power of the radiation the atmosphere absorbs from Earth equals the power of the radiation it emits.
  5. Half of the power radiated by the atmosphere hits Earth. Write an equation that says that the power Earth absorbs from the atmosphere and the Sun equals the power that it emits.
  6. Solve your two equations for the unknown temperature of Earth.
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