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2 years ago

Design an experiment to test the rate at which temperature changes for two different masses (amounts) of water.

Physics

1 answer:

mrs_skeptik [129]2 years ago

3 0

Answer:Experimental Question: How does the amount of a substance affect the rate at which temperature changes?

It depends on the conductivity of the material. If the shift is extreme, the temperature near the heating / cooling source will be similar to the temperature of the heating / cooling source and it will take time for the remainder of the material to rise to temperature. It will depend on the conductivity of the material.

Hypothes is:

Materials List:

• digital stopwatch

• 250ml beaker

• rubber bung

• thermometer

• bunsen burner

• tripod

• gauze

• retort stand and clamp

• goggles

Safety Procedures *:

1. Adult supervision is required.

2. Wear safety goggles, apron, and closed-toe shoes.

3. Do not wear baggy sleeves or dangling jewelry. Tie long hair back.

4. Use hot pads or oven mitts to handle hot objects.

5. Do not reach over a hot burner.

6. Do not leave the experiment unattended.

7. Clean up spills immediately.

8. Report any injuries to your Learning Coach or adult supervisor immediately

Experimental Procedures :

• Fill an empty beaker with exactly 150ml of water (check side-scale of beaker)

• Set up apparatus as shown above. Ensure the thermometer is about 2cm above the bottom of the beaker.

• Light the bunsen burner and put on a blue flame. Heat up the water.

• When the temperature on the thermometer has reached 90°C, immediately switch off the burner.

• Start the stopwatch and time for 5.0 minutes.

• Read the thermometer value at the 5.0 minute mark.

• Before repeating the experiment, check the level of water is still 150ml

Data Table:

Start Temperature of Water (°C) Temperature after 5min (°C) Drop in Temperature

(°C) Average Rate of Cooling x 1000 (°C/s)

80 70 10 17

75 66 9 15

70 62 8 13

65 59 6 10

60 55 5 8

Analysis:

Conclusions : There is a strong correlation between the average rate of cooling and the start temperature: the greater the start temperature, the faster the average rate of cooling.

Explanation:

use quillbot or this will be considered plagerism

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A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is

AveGali [126]

Explanation:

For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.

ΔU=Q−W

We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.

W=FΔx

W=3N×2m

W=6J

Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.

ΔU=Q−W

ΔU=10J−6J

ΔU=4J

Answer is 4J

i think this may help you very much

3 0

2 years ago

A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co

Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

m₁ = 0.973kg

when 1.0kg of aluminium block is used, the final temperature of the mixture will be T

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be

heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T = 418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T = 104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0

3 years ago

a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = 181 kg-m/s

relative to the dock. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = 0.12 m/s , away from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction toward the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = 0.62 m/s away from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from that at 0.62 m/s, and the person standing
on the dock sees you start to move away from him at 0.12 m/s, and that's the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

3 years ago

In order to transmit information via radio waves, the waves need to be changed somehow. For car radios this can happen in two wa

laila [671]

Answer: amplitude

Explanation: This describes the maximum amount of the displacement of a particle from it rest position. Usually, it is measured in metres

Since we are considering AM which is amplitude modulation, a technique used in electronic communication, most commonly for broadcasting information through a radio carrier wave. In amplitude modulation, the amplitude (signal strength) of the carrier wave is diversified in proportion to that of the message signal being broadcasted.

7 0

3 years ago

According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f

Bond [772]

Answer:

$S_{s}=300 m/s$

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

$\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s$

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

$\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km$

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0

3 years ago