For the experiment that you devised
[link] , which variables can be changed, and how should they be changed, so as to shorten the time in which a measurement is made?
Use a smaller quantity of ice (smaller
m ).
Use containers with greater thickness (larger
d ).
Use containers with smaller surface areas (smaller
A ).
Use a lower ambient temperature outside the container (smaller
T2 ).
You wish to design a saucepan that has the same rate of thermal conduction as a pan made of silver. You need to use a less costly material, and limit the size of the pan so that the surface area in contact with a range heating element is no more than 50% greater than that of the hypothetical silver pan. Explain what other factor(s) can be adjusted, and by how much, so that your design will be successful. Use
[link] to obtain thermal conductivity values for different substances.
Heat conduction is the transfer of heat between two objects in direct contact with each other.
The rate of heat transfer
(energy per unit time) is proportional to the temperature difference
and the contact area
and inversely proportional to the distance
between the objects:
Conceptual questions
Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while it is safe to touch the surface only a few centimeters away. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top?
Loose-fitting white clothing covering most of the body is ideal for desert dwellers, both in the hot Sun and during cold evenings. Explain how such clothing is advantageous during both day and night.
(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is
and their inside surface is at
, while their outside surface is at
. (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction?
The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a
window that is
thick (1/4 in) if the temperatures of the inner and outer surfaces are
and
, respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation.
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