CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Conduction Law

Two bodies of...

Question

Two bodies of masses $m_{1}$ and $m_{2}$ and specific heat capacities $s_{1}$ and $s_{2}$ are connected by a rod of length l, cross -sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time t = 0, the temperature of thr first body is $T_{1}$ and the temperature of the second body is $T_{2} (T_{2} > T_{1})$ . Find the temperature difference between the two bodies at time t.

Open in App

Solution

$\frac{Q}{t} = \frac{K A (T_{1} - T_{2})}{L}$

Rise in temperature in

$T_{2} = \frac{K A (T_{1} - T_{2})}{L m s}$

Fall in temperature

$T_{1} = \frac{K A (T_{1} - T_{2})}{L m_{1} s_{1}}$

Final temperature in

$T_{1} - T_{1} - \frac{K A (T_{1} - T_{2})}{L m_{2} s_{2}}$

Change in Temperature

= $T_{1} - \frac{K A (T_{1} - T_{2})}{L m_{1} s_{1}}$

$= (T_{2} + \frac{K A (T_{1} - T_{2})}{L m_{2} s_{2}})$

$= (T_{1} - T_{2})$

$- [\frac{K A (T_{1} - T_{2})}{L m_{1} s_{1}} + \frac{K A (T_{1} - T_{2})}{L m_{2} s_{2}}]$

$l n d T = \frac{K A}{L} (\frac{M_{2} S_{2} + M_{1} S_{1}}{M_{1} S_{1} M_{2} S_{2}})$

So difference in temperature

= $(T_{2} - T_{1}) e^{- λ t}$

where $λ = \frac{K A}{L} \frac{(m_{1} s_{1} + m_{2} s_{2})}{m_{1} s_{1} m_{2} s_{2}}$

Suggest Corrections

thumbs-up

2

Similar questions

Q. Two chunks of metal with heat capacities

C_{1}

C_{2}

are interconnected by a rod of length

l

and cross-sectional area A and fairly low conductivity k. The whole system is thermally insulated from the environment. At a moment t = 0, the temperature difference between two chunks of metal equals

(Δ T)_{0}

. Assuming the heat capacity of the rod to be negligible, find the temperature difference between the chunks as a function of time.

Q. A solid body

X

of heat capacity

C

is kept in an atmosphere whose temperature is

T_{A} = 300 K

t = 0

the temperature of

X

T_{0} = 400 K

. It cools according to Newton's law of cooling. At time

t_{1}

its temperature is found to be

350 K

t_{1}

X

is connected to a larger body

Y

at atmospheric temperature

T_{A}

through conducting rod of length

L

cross-sectional area

A

and thermal conductivity

k

. The heat capacity of

Y

is so large that any variation in its temperature may be neglected. The cross-sectional area

A

of the connecting rod is small compared to the surface area of

X

. Find the temperature of

X

t = 3 t_{1}

.

Q. Two blocks with heat capacities

C_{1}

C_{2}

are connected by a rod of length l, cross-sectional are A and heat conductivity K. Initial temperature difference between the two blocks is

T_{0}

. Assuming the entire system to be isolated from surroundings, heat capacity of the rod to be negligible. The temperature difference between the blocks as a function of time is.

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. Two rods having same length and cross-sectional area are connected individually between temperature

T_{1}

T_{2}

. If the rate of loss of heat due to condution is equal and specific heat of two rods are in the ratio

1 : 2

, then the ratio of their thermal conductivity will be

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Conduction

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon