# A2-quart steel pan, 10' in diameter with a wall thickness of 1.11mm is filled with water

###### Question:

## Answers

Aluminium

Explanation:

If we calculate the heat transfer for all the substances, we have

Mc(∆T)

Where M is mass

C is heat capacity

∆T is temperature change.

For iron

Mc(∆T) = 500 ×0.440×100= 22000J

For copper

Mc(∆T) = 500 ×0.385×100= 19250J

For aluminium

Mc(∆T) = 500 ×0.897×100= 44850J

Aluminium has the highest heat capacity and would take longest for this heat capacity to be dissipated under similar condition.

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The first law of thermodynamics states that the change in the internal energy of a system equals the net heat transfer into the system minus the net work done by the system. This is another way of saying that energy can neither be created nor destroyed, but it can be converted from one form to another form.

A tea kettle that is warmed on an electric stove is receiving heat energy, this heat energy increases the kinetic energy of the water particles in the kettle and makes them to move faster. As time goes on, the water begins to turn into vapors. The heat energy that is released into the system is been used to carry out the work of evaporation and the whistling of the kettle.

Explanation:

The kettle on the stove takes thermal energy from the stove and uses it to convert water into steam at constant temperature. The molecules of water vibrate with increasing speeds when the temperature of water increases. When the temperature of water reaches the boiling point, water starts to change its state to steam.

Explanation:

volume of water in the kettle, V = [tex]774 cm^{3}[/tex]

Given:

Power output of burner, P = 2.0 kW = 2000 W

Mass of kettle, m = 810 g = 0.81 kg

Temperature of water in the kettle, T = [tex]20^{\circ}C[/tex]

Time taken by water to boil, t = 2.4 min = 144 s

Temperaturre at boiling, T' = [tex]100^{\circ}C[/tex]

Solution:

Now, we know that:

Iron's specific heat capacity, [tex]c = 0.45 J/g ^{\circ}C[/tex]

Water's specific heat capacity, [tex]c' = 4.18 J/g ^{\circ}C[/tex]

Now,

Total heat, q = Pt

q = [tex]2000\times 144 = 288 kJ[/tex]

Now,

q = (mc +m'c')(T' - T)

[tex]288\times 10^{3} = (0.81\times 0.45 + m'\times 4.18)(100^{\circ} - 20^{\circ})[/tex]

Solving the above eqn m', we get:

m' = 774 g

Now, the volume of water in the kettle, V:

[tex]V = \frac{m'}{\rho}[/tex]

where

[tex]\rho = density of water = 1 g/cm^{3}[/tex]

Now,

[tex]V = \frac{774}{1}[/tex]

Volume, V = [tex]774 cm^{3}[/tex]

A metal pot is filled with water and placed on top of a stove. As the stove top heats up, the water begins to boil and steam can be seen escaping from thepot. Open system. The matter can evaporate and leave in the form of steam or re condense and enter back into the system.

Based on the scenario, the matter escaping from the pot is an example of an : C. Open system because heat and matter are both able to enter or leave the system

The key word here is 'steam can be seen escaping from the pot' , which indicate that the matter and heat able to leave the system after it's entered.

hope this helps

H

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

Wavelength = pie*69+69 with ur mom