A metal bar is used to conduct heat.When the temperature at one end is 100 $\degree$C and at the other is 20 $\degree$C,heat is transferred at a rate of 16 J/s.If the temperature of the hotter end is reduced to 80 $\degree$C,what will be the rate of heat transfer?

Question

Quizplus · Accepted Answer

We can use the formula for heat transfer through a metal bar:

Q/t = kA(delta T)/L

Where Q/t is the rate of heat transfer, k is the thermal conductivity of the metal, A is the cross-sectional area of the bar, delta T is the temperature difference between the two ends, and L is the length of the bar.

Since k, A, and L are constant, we can set up a proportion:

(Q/t)1/(Q/t)2 = (delta T)1/(delta T)2

where 1 represents the initial conditions and 2 represents the new conditions.

Plugging in the initial values, we have:

16/t = kA(100 - 20)/L

Simplifying, we get:

t = kAL/16(80 - 20)

Plugging in the new temperature and solving for the new rate of heat transfer:

Q/t = kA(80 - 20)/L

Q/t = 3/4(16 J/s)

Q/t = 12 J/s

Therefore, the answer is (D) 12 J/s.

A Metal Bar Is Used to Conduct Heat °\degree° C and at the Other Is 20

A Metal Bar Is Used to Conduct Heat $\degree$ C and at the Other Is 20