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3 years ago

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A quantity of 14.1 cm^3 of water at 8.4°C is placed in a freezer compartment and allowed to freeze to solid ice at -7.2°C. How m

any joules of energy must be withdrawn from the water by the refrigerator?

1 answer:

serious [3.7K]3 years ago

3 0

Answer:920.31 J

Explanation:

Given

Volume of water (V) $=14.1 cm^3$

mass(m) $=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm$

Temperature $=8.4^{\circ} C$

Final Temperature $=-7.2 ^{\circ}C$

specific heat of water(c) $=4.184 J/g-^{\circ}C$

Therefore heat required to removed is

$Q=mc(\Delta T)$

$Q=14.1\times 4.184\times (8.4-(-7.2))$

$Q=920.31 J$

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when an element tends to lose its valence electrons in chemical reactions , does it behave more like a metal or nonmetal

Juli2301 [7.4K]

It behaves more like a metal

Explanation:

When an element tends to lose its valence electrons in chemical reactions, they behave more like a metal.

Metals are electropositive.

Electropositivity or metallicity is the a measure of the tendency of atoms of an element to lose electrons.

This is closely related to ionization energy and the electronegativity of the element.

The lower the ionization energy of an element, the more electropositive or metallic the element is .

Metals are usually large size and prefers to be in reactions where they can easily lose their valence electrons.

When most metals lose their valence electrons, they attain stability.

Non-metals are electronegative. They prefer to gain electrons.

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3 years ago

Heat moves from hot to cold either through air or liquid <br>

KonstantinChe [14]

Answer:

both

Explanation:

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3 years ago

A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the

zysi [14]

As per the question, the mass of meteorite [ m]= 50 kg

The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

$Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2$

$=\frac{1}{2}50kg*[1000\ m/s]^2$

$=\frac{1}{2}50* 10^{6}\ J$

$=25*10^6\ J$

Here, J stands for Joule which is the S.I unit of energy.

$Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J$

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3 years ago

Read 2 more answers

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

$\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}$

Put the value into the formula

$525=\sqrt{\dfrac{c+v}{c-v}}\times950$

$\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2$

$\dfrac{c+v}{c-v}=0.305$

$c+v=0.305\times(c-v)$

$v(1+0.305)=c(0.305-1)$

$v=\dfrac{0.305-1}{1+0.305}c$

$v=−0.532c$

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0

3 years ago

Un camion de envios se encuentra detenido en una señal de pare, permitiendo que pase una ambulancia. Inicia su recorrido y al ca

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Answer:

$0.741\ \text{m/s}^2$

Explanation:

v = Velocidad final = $40\ \text{km/h}=\dfrac{40}{3.6}\ \text{m/s}$

u = Velocidad inicial = 0

t = Tiempo empleado = 15 s

a = Aceleración

De las ecuaciones cinemáticas tenemos

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{40}{3.6}-0}{15}\\\Rightarrow a=0.741\ \text{m/s}^2$

La aceleración del camión en el primer intervalo de tiempo es $0.741\ \text{m/s}^2$ .

4 0

3 years ago