1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
babymother [125]
3 years ago
12

A steel casting weighing 2 kg has an initial temperature of 500°c; 25 kg of water initially at 25°c is contained in a perfectly

insulated steel tank weighing 5 kg. the casting is immersed in the water and the system is allowed to come to equilibrium. what is its final temperature? ignore the effects of expansion or contraction, and assume constant specific heats of 4.18 kj⋅kg−1⋅°c−1 for water and 0.50 kj⋅kg−1⋅°c−1 for steel.
Physics
2 answers:
gayaneshka [121]3 years ago
5 0

Answer:

26.6C

Explanation:

Using an energy balance:

Decrease in internal energy of casting must lead to an increase in internal energy of the tank and water, assuming that no heat flows out of the tank - perfectly insulated.

m_{casting}*C_{p,casting}*(T_{casting} - T_{final})= m_{water}*C_{p,water}*(T_{final} - T_{water}) + m_{tank}*C_{p,tank}*(T_{final} - T_{tank})

2*0.5*(500 - T_{final})= 70*4.18*(T_{final} - 25) + 5*0.5*(T_{final} - 25)\\\\(1+2.5+292.6)*T_{final} = 500+7315+62.5\\\\T_{final} = 26.6C

SCORPION-xisa [38]3 years ago
3 0
<span> Plan: Use Q = m · c · ΔT three times. Hot casting cools ΔT_hot = 500°C - Tf. Cold water and steel tank heat ΔT_cold = Tf - 25°C. Set Q from hot casting cooling = Q from cold tank heating.
here
m_cast · c_steel · ΔT_hot = (m_tank · c_steel + m_water · c_water) · ΔT_cold

m_cast · c_steel · (500°C - Tf) = (m_tank · c_steel + m_water · c_water) · (Tf - 25°C)

2.5 kg · 0.50 kJ/(kg K°) · (500°C - Tf) = (5 kg· 0.50 kJ/(kg K°) + 40 kg· 4.18 kJ/(kg K°)) · (Tf - 25°C)

Solve for Tf, remember that K° = C° (i.e. for ΔT's) </span>
You might be interested in
Which of the following best describes the use of a renewable resource? (2 points) Most power plants burn fossil fuels to generat
kolbaska11 [484]

Which of the following best describes the use of a renewable resource?

Answer:

There are areas in California that use the heat from deep in the Earth to generate electricity.

Explanation:

It comes from the earth itself and we use a lot of things that comes from the earth and deep within it.

5 0
3 years ago
A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density
Yuliya22 [10]

The solution of Sulfuric Acid (H2SO4) has the following mole fractions:

  • mole fraction (H2SO4)= 0.034
  • mole fraction (H2O)= 0.966

To solve this problem the formula and the procedure that we have to use is:

  • n = m / MW
  • = ∑ AWT
  • mole fraction = moles of A component / total moles of solution
  • ρ = m /v

Where:

  • m = mass
  • n = moles
  • MW = molecular weight
  • AWT = atomic weight
  • ρ = density
  • v = volume

Information about the problem:

  • m solute (H2SO4) = 17.75 g
  • v(solution) = 100 ml
  • ρ (solution)= 1.094 g/ml
  • AWT (H)= 1 g/mol
  • AWT (S) = 32 g/mol
  • AWT (O)= 16 g/mol
  • mole fraction(H2SO4) = ?
  • mole fraction(H2O) = ?

We calculate the moles of the H2SO4 and of the H2O from the Pm:

MW = ∑ AWT

MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4

MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)

MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol

MW (H2SO4)=  98 g/mol

MW (H2O)= AWT (H) * 2 + AWT (O)

MW (H2O)= (1 g/mol * 2) + (16 g/mol)

MW (H2O)= 2 g/mol + 16 g/mol

MW (H2O)=  18 g/mol

Having the Pm we calculate the moles of H2SO4:

n = m / MW

n(H2SO4) = m(H2SO4) / MW (H2SO4)

n(H2SO4) = 17.75 g / 98 g/mol

n(H2SO4) = 0.1811 mol

With the density and the volume of the solution we get the mass:

ρ(solution)= m(solution) /v(solution)

m(solution) = v(solution) * ρ(solution)

m(solution) = 100 ml * 1.094 g/ml

m(solution) = 109.4 g

Having the mass of the solution we calculate the mass of the water in the solution:

m(H2O) = m(solution) - m solute (H2SO4)

m(H2O) = 109.4 g - 17.75 g

m(H2O) = 91.65 g

We calculate the moles of H2O:

n = m / MW

n(H2O) = m(H2O) / MW (H2O)

n(H2O) = 91.65 g / 18 g/mol

n(H2O) = 5.092  mol

We calculate the total moles of solution:

total moles of solution = n(H2SO4) + n(H2O)

total moles of solution = 0.1811 mol + 5.092  mol

total moles of solution = 5.2731 mol

With the moles of solution we can calculate the mole fraction of each component:

mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution

mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol

mole fraction (H2SO4)= 0.034

mole fraction (H2O)= moles of (H2O) / total moles of solution

mole fraction (H2O)= 5.092  mol / 5.2731 mol

mole fraction (H2O)= 0.966

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

  • Solvent
  • Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

8 0
1 year ago
calculate the percentage increase in speed of the cyclist when the power output changes from 200W to 300W
Likurg_2 [28]

Answer:

50%

Explanation:

That would be the amount

4 0
2 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
A 3.0-kg cart is rolling across a frictionless, horizontal track toward a 1.3-kg cart that is held initially at rest. The carts
Karolina [17]

Answer:

Explanation:

a ) Momentum of first cart = mass x velocity

= 3 x 4.6 =+13.8 kg m /s

Momentum of second cart = 1.3 x - 1.9 = - 2.47 kg m /s

Total momentum = 13.8 - 2.47

= +11.33 kg m /s

b )

Let the velocity of first cart be v at the moment when second cart was at rest

total momentum = 3 x v + 0 = 3 v

Applying conservation of momentum law

3 v  = +11.33

v = +3.77 m /s

6 0
3 years ago
Other questions:
  • How is the mass number calculated for an element?
    5·1 answer
  • 2 QUESTIONS 20 POINTS
    7·2 answers
  • A mobile travels 98 km in 2h calculate: a) your speed b) how many kilometers will you travel in 3h with the same speed?
    6·1 answer
  • A Danish astronomer who used the diameter of the earth's orbit in his calculation of the speed of light was .
    14·2 answers
  • Name at least three physical properties of the bowling<br> ball.
    15·2 answers
  • Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe
    6·1 answer
  • An oscillator with frequency f = 2.1×10^(12) Hz (about typical for a greenhouse gas molecule) is in equilibrium with a thermal r
    11·1 answer
  • What is the magnitude of the resultant of a 7.0-N force acting vertically upward and a 5.0-N force acting horizontally.
    8·1 answer
  • How could you demomstrate a complete and incomplete loop in an electrical circut
    12·1 answer
  • The lung volume that represents the maximum amount of exchangeable air is the _____
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!