An aluminum-alloy rod has a length of 10.00 cm at 20.000C and a length of 10.015 cm at the boiling point of water.
(a) What is the length of the rod at the freezing point of water?
(b) What is the temperature if the length of the rod is 10.009 cm?
Thermal expansion can occur due to an increase in temperature. It has three types’ i.e. linear expansion, surface expansion, and volume expansion. For the given problem, we have to use the formula for linear expansion to find the length of the rod and the temperature of the rod.
The linear expansion of a body, …(i)
Using equation (i), the coefficient of linear expansion for aluminum-alloy is given as:
We have to calculate the change in length of the rod at freezing point of the water so that the change in temperature is given as:
Length of rod at is
So, the change in linear expansion of the rod is given as:
Total length at 00C is given as the sum of original length and the expanded length, that is:
Hence, the value of the length of the rod is 9.996 cm
The length of the rod is given as
But we know that
From equation (i), the change in the temperature due to expansion is given as:
where, Tx is the temperature at which the length of the rod is
Hence, the value of the required temperature is .
Leidenfrost Effect. A water drop that is slung onto a skillet with a temperature between and about will last about . However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance L in Figure). Let , and assume that the drop is flat with height and bottom face area . Also assume that the skillet has a constant temperature and the drop has a temperature of .Water has density , and the supporting layer has thermal conductivity . (a) At what rate is energy conducted from the skillet to the drop through the drop’s bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?
A 0.400 kg sample is placed in a cooling apparatus that removes energy as heat at a constant rate. Figure 18-32 gives the temperature T of the sample versus time t; the horizontal scale is set by ts=80.0 min. The sample freezes during the energy removal. The specific heat of the sample in its initial liquid phase is 300 J/kgK . (a) What is the sample’s heat of fusion and (b) What is its specific heat in the frozen phase?
Figure 18-56a shows a cylinder containing gas and closed by a movable piston.The cylinder is kept submerged in an ice–water mixture. The piston is quickly pushed down from position 1 to position 2 and then held at position 2 until the gas is again at the temperature of the ice–water mixture; it then is slowly raised back to position 1. Figure 18-56b is a p-V diagram for the process. If of ice is melted during the cycle, how much work has been done on the gas?
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