Question

a cylindrical copper rod of diameter 0.04 and length of 0.5 metre is heated at one end by steam the other end is capital zero degree celsius find the quantity of heat conducted a cross section per minute​

Answer 1

Answer:

Basic equation connecting thermal conductivity extrinsic factors like heat flow, temperature gradient area and length is

Rate of heat flow

=

Thermal Conductivity

×

Temp Difference

×

Area

Length

.

Q

=

k

×

Δ

T

×

A

L

......(1)

Note that

.

Q

is in Watts or Calory per second; it is power, or energy per unit time.

Heat

Q

required to melt

10

grams of ice

=

m

×

L

where

L

is latent heat of fusion of water and is

=

79.7

c

a

l

⋅

g

−

1

Q

=

10

×

79.7

=

797

c

a

l

Thermal conductivity of Copper

k

=

0.99

c

a

l

⋅

s

−

1

c

m

−

1

⋅

K

−

1

Inserting given values in equation (1) we get

.

Q

=

0.99

×

(

100

−

0

)

×

π

×

1

2

120

⇒

.

Q

=

2.59

c

a

l

⋅

s

−

1

Number of seconds required to conduct

797

c

a

l

of heat

=

797

2.59

=

308

s

, rounded to last second

Answer 2

The heat conducted is 1492.8 cal/min.

Explanation:

Given that,

Diameter = 0.04 m

Length = 0.5 m

Temperature = 0°C

We need to calculate the quantity of heat conducted a cross section per minute​

Using formula of heat

$Q=k\times\Delta T\times\dfrac{A}{l}$

Where, l = length

A = area

k = thermal conductivity

$\Delta T$ = change in temperature

Put the value into the formula

$Q=0.99\times(100-0)\times\dfrac{\pi\times(0.02)^2}{50}$

$Q=24.881\ cal/s$

$Q=1492.8\ cal/min$

Hence, The heat conducted is 1492.8 cal/min.

Learn more :

Topic : heat

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