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2 years ago

When an ice cube sits out on a warm day and melts where does the heat go?

Physics

2 answers:

arsen [322]2 years ago

7 0

Here ill explain

There isn't any heat , there is a lack of heat. The sun warms the lack of heat (ice) until its just as warm as everything around it . hoped this helped!

prohojiy [21]2 years ago

7 0

Answer:

When an ice cube sits out on a warm day and melts the heat goes into the ice

Explanation:

When an ice cube sits out on a warm day and melts where does the heat go?

Firstly, temperature is the degree of hotness or coldness of a body, while heat is the transfer of internal energy due to temperature difference.

When ice melts there is no temperature change. The heat needed to change it from its solid state to its liquid state any noticeable change in temperature is called Heat of fusion Q measured in joules.

Q=m $L_{f}$

So the heat goes directly to the ice to change its state from solid to liquid without any change in temperature. Only to break its intermolecular force of attraction.

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Maria feels overwhelmed by everything she needs to do for school by next week. How can she best manage this stress?

FrozenT [24]

Answer: Option (b) is the correct answer.

Explanation:

Stress is a situation in which our mind or body feel the pressure from our surroundings, ideas or thoughts. Stress causes distress in the body and mind of an individual.

For example, Ria has not done her homework and she is worried what punishment might her teacher give her when she get to know this.

When can get relive from stress by talking about our thoughts or worries with an individual. The other person might help us solve our problems or he might suggest us in a better way.

Therefore, we can conclude that Maria best manage this stress by talking to family, friends, and teachers for support.

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2 years ago

Read 2 more answers

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

centripetal acceleration $(a_c)=9\times 10^22 m/s^2$

we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

$v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}$

$v=\sqrt{47.7\times 10^11}$

$v=21.84\times 10^5 m/s$

(b)For n=10

$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

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3 years ago

Do longitude lines run horizontally (east-west) or vertically (north-south)?

nlexa [21]

Longitude- Horizontal (East West)
Latitude- Vertical (North South)

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3 years ago

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

$y = \frac{1}{2} gt^2+0+0$