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3 years ago

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Homes may be heated by pumping hot water through radiators. What mass of water (in g) will provide the same amount of heat when

cooled from 90.7 to 39.4°C, as the heat provided when 182 g of steam is cooled from 118°C to 100.°C. (Assume that the specific heat of liquid water is 4.187 J/g·°C over the given temperature range. Also assume that the specific heat of water vapor above its boiling point is 2.078 J/g·°C.)

1 answer:

Nitella [24]3 years ago

3 0

Answer:

a mass of water required is mw= 1273.26 gr = 1.27376 Kg

Explanation:

Assuming that the steam also gives out latent heat, the heat provided should be same for cooling the hot water than cooling the steam and condense it completely:

Q = mw * cw * ΔTw = ms * cs * ΔTw + ms * L

where m = mass , c= specific heat , ΔT=temperature change, L = latent heat of condensation

therefore

mw = ( ms * cs * ΔTw + ms * L )/ (cw * ΔTw )

replacing values

mw = [182g * 2.078 J/g°C*(118°C-100°C) + 118 g * 2260 J/g ] /[4.187 J/g°C * (90.7°C-39.4°C)] = 1273.26 gr = 1.27376 Kg

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PythonA group of statisticians at a local college has asked you to create a set of functionsthat compute the median and mode of

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Answer:

def median(l):
if(len(l) == 0):
return 0
else:
l.sort()
if(len(l)%2 == 0):
index = int(len(l)/2)
mid = (l[index-1] + l[index]) / 2
else:
mid = l[len(l)//2]
return mid
def mode(l):
if(len(l)==0):
return 0
mode = max(set(l), key=l.count)
return mode
def mean(l):
if(len(l)==0):
return 0
sum = 0
for x in l:
sum += x
mean = sum / len(l)
return mean
lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
print(mean(lst))
print(median(lst))
print(mode(lst))

Explanation:

Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).

In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.

In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).

In the main program, we test the three functions using a sample list and we shall get

20.5

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3 years ago

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Answer:

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Explanation:

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2 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

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3 years ago

Which of the following requires formwork?

vekshin1

Answer:

D

Explanation:

Masonry uses stone work, making a stone wall requires perfect masonry.

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3 years ago