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3 years ago

15

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

rcury in this process?

1 answer:

loris [4]3 years ago

5 0

To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$

Where,

m = mass

$C_p$ = Specific Heat

$\Delta T =$ Change in temperature

Replacing with our values we have that

$m = 0.1g$

$C_p = 139J/Kg\cdot K \rightarrow$ Specific heat of mercury

$\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K$

Replacing

$Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J$

Therefore the heat lost by mercury is 0.09J

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Answer:

how much space it takes up in the world

Explanation:

1) Mass is a measurement of the amount of matter something contains, while Weight is the measurement of the pull of gravity on an object. 2) Mass is measured by using a balance comparing a known amount of matter to an unknown amount of matter. Weight is measured on a scale.

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3 years ago

Each value in nature has a number part, called its____<br> and a dimension, or unit

saul85 [17]

Answer:

<u>Magnitude</u>

Explanation:

Each value in nature has a number part, called its magnitude and a dimension called its unit.

For example,

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If the mass of an object is 8 kg and its momentum is -80 kgm/s, what is its velocity?

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An object's momentum is the product of its mass and its velocity:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

p = -80kg×m/s

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Choice D

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3 years ago

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

$\frac{3*4^{2} }{4}$ = 12N

3) a = $10ms^{-2}$ , r = 10m, v=?

F = $\frac{mv^{2} }{r}$ and F = ma

equating the two equations and cancelling a, we have:

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

making v the subject of formula, we have:

v = $\sqrt{ar}$

= $\sqrt{100}$

= 10 $ms^{-1}$

4) r = 10m, v = $5ms^{-1}$ , a = ?

F = $\frac{mv^{2} }{r}$

F = ma

equating the above equations and making a subject of formula, we get:

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

a = 25/10 = 2.5 $ms^{-2}$

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8) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

F = $(3v)^{2}$

F = 9 $v^{2}$

We can see that the force will be 9X greater than it was.

9) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

10) F = $\frac{mv^{2} }{r}$

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

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3 years ago

Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a

harina [27]

Answer:

$K.E.=\dfrac{3}{2}KT$

Explanation:

Given that

Number of particle =N

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$K.E.=\dfrac{3}{2}KT$

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