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3 years ago

How much heat in joules is required to melt an ice cube with a mass of 18.6 g at 0 °C? The Lf for water is 333 J/g.

Physics

1 answer:

EastWind [94]3 years ago

5 0

Answer:

The heat energy required, Q = 6193.8 J

Explanation:

Given,

The mass of ice cube, m = 18.6 g

The heat of fusion of ice, ΔHₓ = 333 J/g

The heat energy of a substance is equal to the product of the mass and heat of fusion of that substance. It is given by the equation,

<em> Q = m · ΔHₓ joules</em>

Substituting the given values in the above equation

Q = 18.6 g x 333 J/g

= 6193.8 J

Hence, the heat required to melt the ice cube is, Q = 6193.8 J

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what moon phase will occur about 14 days after a first quarter?

garik1379 [7]

Answer:

waxing gibbous is the answer.

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3 years ago

A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W

weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, $P_{gauge \ pressure}$ = 2.95 atm

Atmospheric pressure, $P_{atm}$ = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2$

where;

P₁ = $P_{gauge \ pressure} + P_{atm \ pressure}$

P₂ = $P_{atm}$

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }$

where;

$\rho$ is the density of seawater = 1030 kg/m³

$v_2 = \sqrt{ \frac{2(2.95*101325 \ + \ 1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s$

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0

3 years ago

A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu

seropon [69]

Answer:

$\theta=50\ revolution$

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, $\omega_i=600\ rpm=62.83\ rad/s$

Finally it shuts off, $\omega_f=0$

Time taken, t = 10 s

Let $\alpha$ is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{0-62.83}{10}$

$\alpha =-6.283\ rad/s^2$

Let $\theta$ is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }$

$\theta=\dfrac{-62.83^2}{2\times -6.283}$

$\theta=314.15\ radian$

$\theta=49.99\ revolution$

$\theta=50\ revolution$

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0

3 years ago

Explain three ways you can get home safely, without getting behind the wheel, if there are drugs or alcohol in your system.

Ad libitum [116K]

Call your parent, call a local taxi ,or a sober friend

4 0

3 years ago

An object with a charge of -3.2 uC and a mass of 1.0Ã—10^(-2) kg experiences an upward electric force, due to a uniform electric

USPshnik [31]

Answer:

The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.

Explanation:

The detailed solution can be found in the attachment below.

Thank you for reading and I hope this is helpful to you.

6 0

3 years ago