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Tems11 [23]
3 years ago
10

15 g of lead (specific heat = 0.128 J/g Celsius) at 120 Degrees Celsius is placed on a glacier of ice at 0 degrees Celsius. How

much ice melts? (Lf of ice = 333 J/g)
Physics
1 answer:
Anastaziya [24]3 years ago
4 0

This can be solve using heat balance. The amount of heat the lead released is equal to the amount of heat the ice absorb.

The heat release by the leas = m1*Cp*deltaT

Where m1 is the mass of lead

Cp is the specific heat of lead

Delta is the change of temperature

Heat absorbed by the ice = m2Lf

Where m2 is the mass of ice

Lf is the latent heat of fusion of ice

 

M1*Cp*deltaT = m2Lf

(15g) ( 0.128 j/g C) ( 120 -0 C) = m2(333/g)

Solving for m2

<span>M2 = 0.69 kg of ice melts</span>

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Precisely around 1,800 miles below.
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3 years ago
On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the avera
nirvana33 [79]

Solution :-

Given :

Distance 1 = 30 km

Distance 2 = 70 km

We know that speed = distance/time

and, Average speed = total distance/total time taken

When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour

Average speed = 9distance 1 + distance 2)/(time 1 + time 2)

AS time 2 or t2 is time taken for the second part of the journey of 70 km

⇒ 40 = 100/(1 + t2)

⇒ 40 + 40t2 = 100

⇒ 40t2 = 100 - 40

⇒ 40t2 = 60

⇒ t2 = 60/40

⇒ t2 = 1.5

So, t2 or time taken to travel the second part of the journey is 1.5 hours.

Speed of the second part of the journey = distance 2/time 2

⇒ 70/1.5

⇒ 46.666 km/hr or 46.7 km/hr.

Hence the answer is = 46.666 km/hr or 46.7 km/hr.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

:)

3 0
4 years ago
What is the average acceleration of the particle between 0 seconds and 4 seconds? A. 0 meters/second2 B. 0.04 meters/second2 C.
lisov135 [29]

Answer

D. 0.25 meters/second2

Explanation

The average acceleration is the ratio of change in velocity to the change in time of travel.Taking in this case that the change of velocity is a unit, then Average acceleration is given by;

Aacc=Vf-Vi/Tf-Ti

where Vf=final velocity,Vi=initial velocity' Tf=final time, Ti=initial time

Vf-Vi=1m/s

Tf-Ti=4-0=4seconds

Avacc=1/4=0.25m/s2

6 0
3 years ago
150J of heat energy
arsen [322]

Btu/(lb-°F) J/(g-°C i mean this is the correct answer

6 0
2 years ago
Read 2 more answers
An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move
lions [1.4K]
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
8 0
3 years ago
Read 2 more answers
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