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dem82 [27]
3 years ago
10

You pour 250 g of tea into a Styrofoam cup, initially at 80∘C and stir in a little sugar using a 100-g aluminum 20∘C spoon and l

eave the spoon in the cup. Assume the specific heat of tea is 4180 J/kg⋅∘C and the specific heat of aluminum is 900 J/kg⋅∘C. Part APart complete What is the highest possible temperature of the spoon when you finally take it out of the cup?
Physics
1 answer:
jok3333 [9.3K]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to the conservation of energy and heat transferred in a body.

By definition we know that the heat lost must be equal to the heat gained, ie

Q_g = Q_l

Where,

Q = Heat exchange

The heat exchange is defined as

Q = c_p m \Delta T

Where,

c_p = Specific heat

m = mass

\Delta T= Change in Temperature

Therefore replacing we have that

Q_g = Q_l

c_{p-tea} m \Delta T =  c_{p-al} m \Delta T

Replacing with our values we have that

0.25*4180*(80-T) = 0.1*900*(T-20)

11.61*(80-T) = T-20

T= \frac{948.8}{11.61}

T = 75.24\°C

Therefore the highest possible temperature of the spoon when you finally take it out of the cup is 75.24°C

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Explanation:

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4 0
3 years ago
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips
Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

f = \frac{v}{\lambda}

Then the relation between two different frequencies with same wavelength would be

\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}

\frac{f'}{f} = \frac{v'}{v}

The beat frequency heard when the two strings are sounded simultaneously is

f_{beat} = f-f'

f_{beat} = f(1-\frac{f'}{f})

f_{beat} = f(1-\frac{v'}{v})

We have the velocity of the transverse waves in stretched string as

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{200N}{\mu}}

And,

v' = \sqrt{\frac{196N}{\mu}}

Therefore the relation between the two is,

\frac{v'}{v} = \sqrt{\frac{192}{200}}

\frac{v'}{v} = \sqrt{0.96}

Finally substituting this value at the frequency beat equation we have

f_{beat} = 590(1-\sqrt{0-96})

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4 0
2 years ago
What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.
Stolb23 [73]

Answer:

Momentum(p) = 20kgm/s

Explanation:

Given

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Required

Calculate the momentum of the man

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or

p = mv

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

Momentum(p) = 10kg * 2m/s

Momentum(p) = 10 * 2 * kg * m/s

Momentum(p) = 20kgm/s

Hence, the momentum of the man is 20kgm/s

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3 years ago
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krok68 [10]

There's supposed to be a blank in the statement, where the answer is supposed to be inserted.

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