Question #cc588

1 Answer
Jun 23, 2015

Answer:

The one with the greatest #r^2/l# ratio.

Explanation:

Heat transferred by conduction an be calculated using the equation

#Q/t = (k * A * (T_"hot" - T_"cold"))/d#, where

#Q# - heat transferred;
#t# - the time needed to transfer #Q# heat;
#k# - the thermal conductivity of the material;
#A# - the surface of the material;
#d# - the thickness of the material;
#T_"hot"#, #T_"cold"# - the higher temperature and lower temperature, respectively.

So, you know that your rods are built from the same material, which means that #k# is the same for all four rods. Moreover, the temperature difference between the reservoirs is the same for all four rods.

Assuming that you measure the heat transferred in the same time interval for all four rods, you can write

#Q = underbrace(k * (T_"hot" - T_"cold") * t)_(color(blue)("constant")) * A/d#

Therefore,

#Qprop A/d#

So, in your case, the rods are cylinders with radius #r# and length #l#. This means that the area and the thickness of a rod will be

#A = pi * r^2#
#d = l#

This means that you have

#Q prop (pi * r^2)/l <=> Q prop r^2/l# #-># since #pi# is constant as well.

Therefore, the rod that has the largest #r^2/l# ratio will conduct more heat, given that all other parameters are identical for all the four rods.