MyQuestionIcon

3

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Heat

A hot object ...

Question

A hot object with a temperature $T_{1}$ is connected to a cool object with a temperature of $T_{2}$ . The object used to conduct heat has a length L and a cross-sectional area A. The rate of heat flow is :

A

A (T_{1} - T_{2}) / k L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

k (T_{1} - T_{2}) / A L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

k A L / (T_{1} - T_{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

k A (T_{1} - T_{2}) / L

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $k A (T_{1} - T_{2}) / L$
Rate of heat flow is given by
$\frac{Q}{L} = \frac{k A (T_{1} - T_{2})}{L}$ (for heat flow through conduction)
where $k$ = thermal conductivity of material
$A$ = cross-selectional area
$L$ = Length
$△ T = T_{1} - T_{2} =$ change in temperature

Suggest Corrections

thumbs-up

0

Similar questions

Q. A rod of length

l

and cross section area A has a variable thermal conductivity given by

k = α T

α

is a positive constant and

T

is temperature in Kelvin. Two ends of the rod are maintained at temperature

T_{1}

T_{2}

(T_{1} > T_{2})

. Heat current flowing through the rod will be

Q. A thin copper wire length L increases its length by 1% heated from temperature

T_{1}

T_{2}

What is the percentage change in area when a thin copper plate having dimension 2Lx L is heated from

T_{1}

T_{2}

Q. A thin copper wire of length

L

increase in length by

1 %

when heated from temperature

T_{1}

T_{2}

. What is the percentage change in area when a thin copper plate having dimensions

2 L \times L

T_{1}

T_{2}

?

Q. If the sides of a wall having thermal conductivity

K

are maintained at temperatures of

T_{1}

T_{2}

, the temperature as a function of

x

can be written as

T (x) = (T_{2} - T_{1}) (\frac{x}{L}) + T_{1}

Find the heat flux through the wall.
Take

T_{1} > T_{2}

Q. A carnot engine operating between temperature

T_{1}

T_{2}

has the efficiency of

\frac{1}{6}

. When the temperature

T_{2}

is lowered by 62K, it efficiency becomes

\frac{1}{3}

, Then temperature

T_{1}

T_{2}

respectively are

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Heat_tackle

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard VII Physics

Join BYJU'S Learning Program

CrossIcon