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Oduvanchick [21]

3 years ago

10

An 8.5 g ice cube is placed into 255 g of water. Calculate the temperature change in the water upon the complete melting of the

ice. Assume that all of the energy required to melt the ice comes from the water.

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

Change in temperature of water = 2.67 °C

Explanation:

Heat gained by ice = heat lost by water

Q₁ = Q₂............... Equation 1

Where Q₁ = Latent heat of ice, Q₂ = heat lost by water

Q₁ = lm₁.................... equation 2

Q₂ = cm₂ΔT ............... Equation 3

Substituting equation 2 and 3 into equation 1

lm₁ = cm₂ΔT............... Equation 4

Making ΔT the subject of formula in equation 4

ΔT = lm₁/cm₂............. Equation 5

Where ΔT = change in temperature of water, l = specific latent heat of ice, m₁ = mass of ice, c= specific heat capacity of water, m₂ = mass of water.

Given: m₁ = 8.5 g = 0.0085 kg, m₂ = 255 g = 0.255 kg.

Constant: l = 336000 J/kg, c = 4200 J/kg.K

Substituting these values into equation 5

ΔT = (336000×0.0085)/(0.255×4200)

ΔT = 2856/1071

ΔT = 2.67 °C

Change in temperature of water = 2.67 °C

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Komok [63]

Question:

If you push a bowling ball and a golf ball with an equal force what will happen

Answer:

B)

Explanation:

Larger than the force used to push the object that has less mass. A golf ball and a bowling ball are moving at the same velocity. When gravity and air resistance are equal, the object has drawn its terminal velocity.

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Can a physical change, change what a substance is

Elden [556K]

No the substance will remain the same substance as before.

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3 years ago

Is it possible to have a charge of 5 x 10-20 C? Why?

ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is $2.3\cdot 10^{39}$ times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

$e=1.6\cdot 10^{-19}C$

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than $e$ , because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

$Q=5\cdot 10^{-20}C$

If we compare this value to $e$ , we notice that $Q$ , so no object can have a charge of $Q$ .

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

$Q=ne$

where

Q is the charge of the object

n is an integer multiple

e is the fundamental charge

Here we have

$Q=2.4\cdot 10^{-18}C$

Substituting the value of e, we find n:

$n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15$

n is integer, so this value of the charge is possible.

3)

We now do the same procedure for the new object in this part, which has a charge of

$Q=2.0\cdot 10^{-19}C$

Again, the charge on this object can be written as

$Q=ne$

where

n is the number of electrons in the object

Using the value of the fundamental charge,

$e=1.6\cdot 10^{-19}C$

We find:

$n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25$

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

4)

To solve this part, we use Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (Action force), then object B exerts an equal and opposite force on object A (reaction force)".

In this problem, we have two objects:

- A charge Q

- A charge 5Q

Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

5)

The magnitude of the electric force between two single-point charges is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

r = 0.30 cm = 0.003 m is the separation

So, the electric force between the two charges is

$F=(9\cdot 10^9)\frac{(4.5\cdot 10^{-6})(7.2\cdot 10^{-6})}{(0.003)^2}=32400 N$

6)

The electric force between two charged objects has direction as follows:

- If the two objects have charges of opposite signs (+ and -), the force between them is attractive

- If the two objects have charges of same sign (++ or --), the force between them is repulsive

In this problem, the two charges are:

$q_1=+4.5\cdot 10^{-6}C$ is charge 1

$q_2=+7.2\cdot 10^{-6}C$ is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

$F_E=k\frac{q_1 q_2}{r^2}$

where

$q_1 = q_2 = e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the proton and of the electron

$r=5.3\cdot 10^{-11} m$ is the separation between them

So the force can be rewritten as

$F_E=\frac{ke^2}{r^2}$

The gravitational force between the proton and the electron can be written as

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p = 1.67\cdot 10^{-27}kg$ is the proton mass

$m_e=9.11\cdot 10^{-27}kg$ is the electron mass

Comparing the 2 forces,

$\frac{F_E}{F_G}=\frac{ke^2}{Gm_p m_e}=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{(6.67\cdot 10^{-11})(1.67\cdot 10^{-27})(9.11\cdot 10^{-31})}=2.3\cdot 10^{39}$

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Answer

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Explanation:

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8 0

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Read 2 more answers

You and a highway patrolman are driving at constant speeds in opposite directions on a straight highway. The patrolman is drivin

kompoz [17]

Answer: 75 mph

Explanation:

The Relative Speed for a mobile is equal to the diference between the object and the observer:

Relative Speed (Rs) = Object's Velocity - Observer's Velocity

Thinking on those terms, we would need to have a universal observer to do any understandable measurement on daily basics. This is why we all use earth as a Static Observer for every measurement we do everyday.

Using Earth as an observer, the Velocity for the Patrolman is:

Patrolman Velocity (Vp) = 60 mph

Because the radar gun does measure the Relative Speed for the object, which is 135 mph, we need to work with the equation to find the Velocity using Earth as a reference.

Object's Relative Velocity = Object's Velocity - Patrolman's Velocity

Object's Velocity = Object's Relative Velocity + Patrolman's Velocity

We need to keep in mind, the Patrolman is going on the opposite direction. Because of this the sign for his velocity should be negative.

Object's Velocity = 135 mph + ( -60 mph)

Object's Velocity = 135 mph - 60 mph

Object's Velocity = 75 mph

3 0

3 years ago