Ask question

Ask question

All categories

2 years ago

9

How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice

cube? = 6.02 kJ/mol 108 kJ 102 kJ 5.64 kJ 936 J none of the above

1 answer:

Vilka [71]2 years ago

5 0

Answer:

How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice cube? = 6.02 kJ/mol

A. 108 kJ

B. 102 kJ

C. 5.64 kJ

D. 936 kJ

E. none of the above

<em>5.64 kJ</em>

Explanation:

The Heat of fusion is the heat energy required to dissolve a given mass of ice at melting point.

<h3>Step by Step Calculation</h3>

The heat energy required to dissolve ice can be calculated using the expression below;

Q = ΔH $_{f}$ x m ...............................................1

where Q is the heat energy required;

ΔH $_{f}$ is the heat of fusion for ice;

m is the mole

All the parameters above are provided in the question except m, so to get m we use the molar mass of water (also for ice) which is 18.01528 g/mol .

<em>This means that 18.01528 g of ice is contained in one mole, therefore the mole for 16.87 g of ice is given as;</em>

$m = \frac{16.87g}{18.015g/mol}$

m = 0.9364 mole of ices

Now the parameters are complete, we are given;

ΔH $_{f}$ = 6.02 kJ/mol

m = 0.9364 mol

Q =?

Substituting into equation 1, we have

Q = 6.02 kJ/mol x 0.9364 mol

Q = 5.64 kJ

<em>Therefore, the energy required to melt 16.87 g of ice is 5.64 kJ</em>

You might be interested in

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

USPshnik [31]

No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.

7 0

2 years ago

Read 2 more answers

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

work done by the gravity = ?

work done by the tension = ?

Work done by the gravity = - m g Δh

Δ h = 2 + 2 = 4 m

Work done by the gravity = $- 3 \times 9.8 \times 4$

= -117.6 J

work done by gravity is equal to -117.6 J

Work done by tension will be equal to zero.

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J

7 0

2 years ago

A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at

Stells [14]

The correct answer is <span>3) $K_f = \frac{1}{2}mv_0^2 + mgh$ .
</span>
In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
</span> $E=U_i+K_i=mgh + \frac{1}{2}mv_0^2$
<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
</span> $E=K_f$ <span>
But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
</span> $K_f = mgh + \frac{1}{2}mv_0^2$ <span>
</span>

7 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago

A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If

sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)= $50 \Omega$

Voltage(V)=15 V

Time constant( $\tau$ )= $\frac{L}{R}$

$\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}$

current $i_0=\frac{V}{R}$

$i_0=\frac{15}{50}=0.3 A$

Current as a function of time is given by

$i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )$

$i=0.3\times 0.9998$

i= 299.95 mA

6 0

2 years ago