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strojnjashka [21]
2 years ago
12

How r u? Are u good?

Physics
2 answers:
meriva2 years ago
8 0
Hi i’m great thanks for asking wbu
Maslowich2 years ago
6 0

Answer:

i'm good, hru hope ur staying safe

Explanation:

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A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0
3 years ago
The process of changing the energy of a system by means of force. Force * Diatance
Dmitry_Shevchenko [17]
This is known as work
7 0
3 years ago
Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)
Ostrovityanka [42]

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

3 0
2 years ago
A train travels 160 km in 2 h. What is the train’s average speed in km/h?
loris [4]
The train’s average speed is 80km/h
3 0
3 years ago
Read 2 more answers
If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?
steposvetlana [31]

Hi there!

\large\boxed{\text{B) 20 meters}}

We know that:

E_T = U + K

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

3 0
2 years ago
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