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elena-14-01-66 [18.8K]
3 years ago
8

A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which

is thermally well insulated except for its contact with the copper. The length of the rod between the containers is 19.6 cm, and the rod is covered with a thermal insulator to prevent heat loss from the sides.
How many grams of ice melt each second? (The thermal conductivity of copper is 390 W/m-C°.)
Physics
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

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If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i
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Answer:

9 cm.

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using the energy of spring formula ,we find that

27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))

45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))

Dividing both the equation will get,

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to given data

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solution

current flow in series is express as here

current = voltage / resistor    .................1

put here all value  in equation 1

current = 5 / (1.5 + 0.5)

current = 5 / 2.0

so current = 2.50 mA

and

current flow in parallel is express as

current = voltage / resistor   ....................2

put here all value in equation  2

current = 5 / (1/ (1/1.5 + 1/0.5))

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