Put a few spoonfuls of water into a bottle with a tight lid. Make sure everything is at room temperature, measuring the temperature of the water with a thermometer to make sure. Now close the bottle and shake it as hard as you can for several minutes. When you're exhausted and ready to drop, shake it for several minutes more. Then measure the temperature again. Make a rough calculation of the expected temperature change, and compare.


Short Answer

Expert verified

Temperature will change due to kinetic energy of molecules.

Step by step solution

01

Given information

Few spoonfuls of water into a bottle with a tight lid.

02

Explanation

Temperature can't change (neither increased nor decreased) without any external factors (heat supply or heat absorption) . Temperature is related to the kinetic energy of the molecules.

When the bottle is kept in the room temperature, then its temperature will be say 25oC.

And when the bottle is shaken for couple of minutes, then its temperature is measured with thermometer. In this case, temperature will be increased due to the kinetic energy of molecules increases during shaking. This change in kinetic energy of molecule causes a change in temperature.


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Most popular questions from this chapter

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 550,000 . The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β :
βΔV/VΔT
(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β =1 / 550,000 K-1=1.81 x 10-4 K-1. (The exact value varies with temperature, but between 0oC and 200oC the variation is less than 1 %.)
(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.
(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10 -4 K-1 at 100oC, but decreases as the temperature is lowered until it becomes zero at 4oC. Below 4oC it is slightly negative, reaching a value of -0.68 x 10-4K-1 at 0oC. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.


Suppose you open a bottle of perfume at one end of a room. Very roughly, how much time would pass before a person at the other end of the room could smell the perfume, if diffusion were the only transport mechanism? Do you think diffusion is the dominant transport mechanism in this situation?

The enthalpy of combustion of a gallon (3.8 liters) of gasoline is about 31,000kcal. The enthalpy of combustion of an ounce28g of corn flakes is about100kcal. Compare the cost of gasoline to the cost of corn flakes, per calorie.

In analogy with the thermal conductivity, derive an approximate formula for the diffusion coefficient of an ideal gas in terms of the mean free path and the average thermal speed. Evaluate your formula numerically for air at room temperature and atmospheric pressure, and compare to the experimental value quoted in the text. How does D depend on T, at fixed pressure?

Imagine some helium in a cylinder with an initial volume of 1 liter and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in direct proportion to its volume.

  1. Sketch a graph of pressure vs. volume for this process.
  2. Calculate the work done on the gas during this process, assuming that there are no “other” types of work being done.
  3. Calculate the change in the helium’s energy content during this process.
  4. Calculate the amount of heat added to or removed from the helium during this process.
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