Question

Ans: 425
A metal Cylinder 0.628m long and 0.04m in diameter has one end in boiling water at 100°C and
other end in melting ice. The Coefficient of Thermal Conductivity of the metal is 378wm 'k.
Latent Hat of Ice is 3.36x10ʻJ/Kg. Find the mass of Ice melts in one hour.
​

Answer 1

Answer:

The mass of ice that melts in one hour is 0.810 kg

Explanation:

As we know that the heat transferred by thermal conduction is given as

$\frac{dQ}{dt} = \frac{kA\Delta T}{L}$

here the heat is used to melt the ice so we will have

$\frac{dQ}{dt} = L\frac{dm}{dt}$

$L\frac{dm}{dt} = \frac{kA\Delta T}{L}$

$(3.36 \times 10^5)(\frac{m}{3600}) = \frac{(378)(\pi (0.02)^2)(100 - 0)}{0.628}$

so we will have

$93.33 m = 75.64$

$m = 0.810 kg$

#Learn

Topic : Thermal conduction

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Answer 2

Answer:

Explanation:ANSWER

l=0.628\ meterl=0.628 meter

A=\pi r^2=\pi \dfrac{d^2}{r}=\dfrac{\pi}{4}\times (0.04)^2=4\pi \times 10^{-4}\ m^2A=πr

2

=π

r

d

2

​

=

4

π

​

×(0.04)

2

=4π×10

−4

m

2

\Delta \theta=100-0=100^o C=373\ k-273\ k=100\ kΔθ=100−0=100

o

C=373 k−273 k=100 k

k=378\dfrac{W}{m\ Kelvin}k=378

m Kelvin

W

​

L=3.36\times 10^5\ J/kgL=3.36×10

5

J/kg

one hour t=60\times 60\ second= 3600\ sect=60×60 second=3600 sec

so heat fastered in tt second will be

Q=kA\dfrac{\Delta Q}{l}.t=378\times \dfrac{4\pi \times 10^{-4}\times 100}{0.628}\times 3600Q=kA

l

ΔQ

​

.t=378×

0.628

4π×10

−4

×100

​

×3600

or Q=2.7216\times 10^9\ JouleQ=2.7216×10

9

Joule

ice melt \boxed{m=\dfrac{Q}{L}=\dfrac{2.7216\times 10^9}{3.36\times 10^5}=8.1\times 10^3\ kg}

m=

L

Q

​

=

3.36×10

5

2.7216×10

9

​

=8.1×10

3

kg

​