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2 years ago

What is the difference between heat and temperature

Physics

1 answer:

Tasya [4]2 years ago

7 0

Temperature is the measurement system we use to measure the heat

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Communicating results allows scientists to learn from each other's results.

sveta [45]

True, scientists often talk to each other to figure out if their results were similar and what they could have done better.

Although, talking to other scientists does have risks, other scientists could copy your work and further better it.

So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol

3 0

2 years ago

Read 2 more answers

If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc

Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0

2 years ago

Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the

Hoochie [10]

You've told us:

-- 130°x = 212°F

and

-- 10°x = 32°F

Thank you. Those are two points on a graph of °x vs °F . With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are (130, 212) and (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation. I'll use the second point (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are ! Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad. Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for x, and the solution will be the same temperature in F too.

We can write [ F = 1.5F + 17 ], solve it for F, and the solution will be the same temperature in x too.

F = 1.5F + 17

Subtract F from each side: 0.5F + 17 = 0

Subtract 17 from each side: 0.5F = -17

Multiply each side by 2 : F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for x:

F = 1.5(-34) + 17

F = -51 + 17

It works ! -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

6 0

2 years ago

A diagram of a closed circuit with a power source on the left labeled 120 V. There are 3 resistors in parallel, separate paths,

Zina [86]

1) The equivalent resistance is $2.73\Omega$

2) The voltage in the circuit is 120 V

3) The total current in the circuit is 44.0 A

Explanation:

Resistors are said to be in parallel when they have the same potential difference at their terminals.

The formula to calculate the equivalent resistance for three resistors in parallel is:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

where $R_1, R_2, R_3$ are the resistances of the three resistors.

In this problem, the resistance of each resistor is:

$R_1 = 5.0 \Omega$

$R_2 = 10.0 \Omega$

$R_3 = 15.0 \Omega$

Substituting,

$\frac{1}{R_T}=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}=\frac{11}{30}$

So the equivalent resistance is

$R_T = \frac{30}{11}=2.73 \Omega$

In this problem we are told that the three resistors are connected in parallel. This means that their terminals are connected to the same point of the circuit: therefore, this means that they also have the same potential difference across them.

Therefore, the voltage in each branch containing each resistor is the same, and it is equal to the voltage of the battery, which is

$V=120 V$