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MrRa [10]
2 years ago
8

The additional energy in a beaker of hot water compared to an otherwise identical beaker of room temperature water is _ _ _ _ _

_ __. Potential energy work thermal energy chemical energy.
Physics
2 answers:
Fynjy0 [20]2 years ago
7 0
<span>The additional energy in a beaker of hot water compared to an otherwise identical beaker of room temperature water is thermal energy.
hope this helps</span>
il63 [147K]2 years ago
6 0

Correct answer choice is :


C) Thermal energy


Explanation:


Thermal energy is a term applied loosely as an analog for more rigorously specified thermodynamic measures such as the physical energy of a system heat or tangible heat, which are described as examples of change of energy or for the specific energy of a degree of choice in a thermal system. Heat energy that is converted into light can be obtained in many ways. It can be generated by heating fuels such as coal, oil, gas or wood. It can also be obtained from steam from a geothermal field or produced by nuclear effects.

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A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg 
Aleonysh [2.5K]

Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

Q = 45 * 334 * 10^3

Q = 1.5 * 10^7 J

In KJ we can convert this as

Q = 1.5 * 10^4 kJ

so the correct answer is D option

7 0
2 years ago
Read 2 more answers
Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC
scoray [572]

Answer:

m = 4.65 kg

Explanation:

As we know that the mass of the water that evaporated out is given as

m = 0.0250 kg

so the energy released in form of vapor is given as

Q = mL

Q = (0.0250)(2.25 \times 10^6)

Q = 56511 J

now the heat required by remaining water to bring it from 15 degree to 100 degree

Q_2 = ms\Delta T

Q_2 = (4 - 0.025)(4186)(100 - 15)

Q_2 = 1.41\times 10^6J

total heat required for above conversion

Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J

now by heat energy balance

heat given by granite = heat absorbed by water

m(790)(500 - 100) = 1.47 \times 10^6

m = 4.65 kg

4 0
2 years ago
A substance has a heat of vaporization of 16.69 kJ/mole. At 254.3 K it has a vaporpressure of 92.44 mm Hg. Calculate its vapor p
adoni [48]

Answer:

The vapor pressure is 170.6 mmHg.

Explanation:

Given that,

Heat of vaporization = 16.69 kJ/mole

Temperature = 254.3

Pressure = 92.44 mmHg

Temperature = 275.7 K

We need to calculate the vapor pressure

Using relation pressure and temperature

ln\dfrac{P_{2}}{P_{1}}=\dfrac{-\Delta H}{R}(\dfrac{1}{T_{2}}-\dfrac{1}{T_{1}})

Put the value into the relation

ln\dfrac{P_{2}}{92.44}=\dfrac{-16690}{8.314}(\dfrac{1}{275.7}-\dfrac{1}{254.3})

ln\dfrac{P_{2}}{92.44}=0.61274

\dfrac{P_{2}}{92.44}=e^{0.61274}

P_{2}=1.84548\times92.44

P_{2}=170.6\ mmHg

Hence, The vapor pressure is 170.6 mmHg.

5 0
2 years ago
Un lector de DVD, la velocidad de giro es de 5400 rpm. determina el valor velocidad angular en rad/s,la frecuencia y el periodo
zubka84 [21]

Responder:

A) ω = 565.56 rad / seg

B) f = 90Hz

C) 0.011111s

Explicación:

Dado que:

Velocidad = 5400 rpm (revolución por minuto)

La velocidad angular (ω) = 2πf

Donde f = frecuencia

ω = 5400 rev / minuto

1 minuto = 60 segundos

2πrad = I revolución

Por lo tanto,

ω = 5400 * (rev / min) * (1 min / 60s) * (2πrad / 1 rev)

ω = (5400 * 2πrad) / 60 s

ω = 10800πrad / 60 s

ω = 180πrad / seg

ω = 565.56 rad / seg

SI)

Dado que :

ω = 2πf

donde f = frecuencia, ω = velocidad angular en rad / s

f = ω / 2π

f = 565.56 / 2π

f = 90.011669

f = 90 Hz

C) Periodo (T)

Recordar T = 1 / f

Por lo tanto,

T = 1/90

T = 0.0111111s

3 0
2 years ago
Mr. Smith is designing a race where velocity will be measured. Which course would allow velocity to accurately get a winner?
liraira [26]
I’m not completely sure but most likely is is the 10 mile bike ride, I hope I can help! (:
6 0
2 years ago
Read 2 more answers
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