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morpeh [17]
3 years ago
8

A 500 W immersion heater is placed in a pot containing 1.00 L of water at 20oC. (a) How long will the water take to rise to the

boiling temperature, assuming that 70.0% of the available energy is absorbed by the water
Physics
1 answer:
tatiyna3 years ago
6 0

Answer:

96 s.

Explanation:

(a)

From the question,

Q = cm(t₂-t₁)................... Equation 1

Where Q = heat required to boil the water, c = specific heat capacity of the water, m = mass of the water, t₂ = final temperature of water, t₁ = initial temperature of water

Note: The boiling point of water = 100 °C

Given: c = 4200 J/kg.°C, t₂ = 100 °C, t₁ = 20 °C

mass of water = density×volume

m = D×v, Where D = 1000 kg/m³, v = 1.00 L = 0.001 m³

Hence, m = 1000×0.001 = 1 kg.

Substitute into equation 1

Q = 4200×1(100-20)

Q = 4200×8

Q = 33600 J.

But,

P = Q/t................... Equation 2

make t  the subject of the equation

t = Q/P................. Equation 3

Where P = power, t = time

From the question,

70 % of the available energy is absorbed by water.

P = 0.7×500 = 350 W.

Substitute into equation 2

t = 33600/350

t = 96 s.

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
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v_e=\sqrt{\frac{2GM}{R}}

where

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M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

v=0.421 v_e

So its initial kinetic energy will be

K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:0.177 \frac{GMm}{R}=\frac{GMm}{nR}

And solving the equation we find

n=\frac{1}{0.177}=5.65

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

This time, the projectile has 0.421 times this energy:

K=0.421 \frac{GMm}{R}

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

0.421 \frac{GMm}{R}=\frac{GMm}{nR}

and by solving for n we find

n=\frac{1}{0.421}=2.36

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) E=U=\frac{GMm}{R}

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

E=U=\frac{GMm}{R}

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

6 0
3 years ago
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