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3 years ago

Dimension of specific heat capacity

Physics

1 answer:

Luden [163]3 years ago

3 0

Answer:

The specific heat of a substance (per unit of mass) has dimension L. T. Θ−1, or (L/T)2/Θ. Therefore, the SI unit J/K/kg is equivalent to metre squared per second squared per kelvin (m2 K−1 s−2).

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Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.

Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, $\Delta T$ , T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

$(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}$

$k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}$

$\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}$

$\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}$

$\Delta T_{air} = 44.9 \Delta T_{glass}$

There are two layers of Glass and one layer of Air so the total temperature would be given as,

$\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}$

$\Delta T = 2\Delta T_{glass} +\Delta T_{air}$

$20\°C = 46.9\Delta T_{glass}$

$\Delta T_{glass} = 0.426\°C$

Finally the rate of heat flow through this windows is given as,

$\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}$

$\Delta {Q}{t} = 0.84*24*10 -3*0.426$

$\Delta {Q}{t} = 179W$

Therefore the correct answer is D. 180W.

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3 years ago

A basketball is thrown straight up in the air. At its peak, it is 0.02 km high and has a velocity of 0 m/s. What is its final ve

Aleks04 [339]

Answer:

20 m/s

Explanation:

Given:

Δy = 0.02 km = 20 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (20 m)

v = 19.8 m/s

Rounded to one significant figure, the final velocity is 20 m/s towards the ground.

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4 years ago

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vichka [17]

What is the definition of newton

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3 years ago

The earth's radius is 6.37×10^6m; it rotates once every 24 hours. with the angular speed of 7.3 x 10^-5 what is the speed of a p

alexira [117]

The speed of the earth's surface located at 2/3 of the length of the arc between the pole which measure from the equator is 232.5 m/s.

Solution:
So the givens are, earth's radius = 6.37X10^6m, and the angular distance from the pole is 90 degrees. So 60 degrees is the 2/3.

r = 6.37x10^6 * cos(60) = 3.185x10^6m
since v = wr
v = 7.3x10^-5 * 3.185x10^6

v - 232.5 m/s

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4 years ago

If you double the mass but don't change the net force what is the result on the acceleration

vazorg [7]

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3 years ago

Dimension of specific heat capacity​

Dimension of specific heat capacity