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kirill [66]
3 years ago
9

What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?

Physics
1 answer:
Effectus [21]3 years ago
5 0

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

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A 10-cm long solenoid has 100 turns and a radius of 5 cm. If it carries a current of 2 A, What is the magnetic field B inside th
OLga [1]

Hi there!

We can use Ampère's Law:

\oint B \cdot dl = \mu_0 i_{encl}

B = Magnetic field strength (B)
dl = differential length element (m)
μ₀ = Permeability of free space (T/Am)

Since this is a closed-loop integral, we must integrate over a closed loop. We can integrate over a rectangular-enclosed area of the rim of the solenoid - ABCD - where AD and BC are perpendicular to the solenoid.

Thus, the magnetic field is equivalent to:
\oint B \cdot dl = \int\limits^A_B {B} \, dl  + \int\limits^B_C {B} \, dl   + \int\limits^C_D {B} \, dl   + \int\limits^D_A {B} \, dl

Since AD and BC are perpendicular, and since:
\oint B \cdot dl = B \cdot L = BLcos\phi

BLcos(90) = 0

If perpendicular to the field, the equation equals 0.

Additionally, since AB is outside of the solenoid, there is no magnetic field present, so B = 0. The only integral we integrate now is:
\oint B \cdot dl = \int\limits^C_D {B} \, dl

Which is horizontal and inside the solenoid. Let the distance between C and D be 'L', and the enclosed current is equivalent to the number of loops multiplied by the current:

B L = \mu_0 Ni

N = # of loops per length multiplied by the length, so:
BL = \mu_0 nL i \\\\B = \mu_0ni

Plug in the given values and solve. Remember to convert # of loops to # of loops per unit length.

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6 0
2 years ago
How many kilojoules is this?
Margaret [11]

Answer:

How to convert joules to kilojoules

One joule is equal to thousandth of a kilojoule:

1J = 0.001kJ

The energy in kilojoules E(kJ) is equal to the energy in joules E(J) divided by 1000:

E(kJ) = E(J) / 1000

Example

Convert 500 joules to kilojoules.

The energy E in kilojoules (kJ) is equal to 500 joules (J) divided by 1000:

E(kJ) = 500J / 1000 = 0.5kJ

Joules to kilojoules conversion table

Energy (J) Energy (kJ)

1 J 0.001 kJ

2 J 0.002 kJ

3 J 0.003 kJ

4 J 0.004 kJ

5 J 0.005 kJ

6 J 0.006 kJ

7 J 0.007 kJ

8 J 0.008 kJ

9 J 0.009 kJ

10 J 0.01 kJ

20 J 0.02 kJ

30 J 0.03 kJ

40 J 0.04 kJ

50 J 0.05 kJ

60 J 0.06 kJ

70 J 0.07 kJ

80 J 0.08 kJ

90 J 0.09 kJ

100 J 0.1 kJ

200 J 0.2 kJ

300 J 0.3 kJ

400 J 0.4 kJ

500 J 0.5 kJ

600 J 0.6 kJ

700 J 0.7 kJ

800 J 0.8 kJ

900 J 0.9 kJ

1000 J 1 kJ

2000 J 2 kJ

3000 J 3 kJ

4000 J 4 kJ

5000 J 5 kJ

6000 J 6 kJ

7000 J 7 kJ

8000 J 8 kJ

9000 J 9 kJ

10000 J 10 kJ

100000 J 100 kJ

hope it help u

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Answer:

60 m

Explanation:

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