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2 years ago

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The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r

aise the temperature of 0.900 kg of this sea water from 10 °C up to its boiling point. Also mention the equation to be used. *

1 answer:

maw [93]2 years ago

6 0

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

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loris [4]

Questions Diagram is attached below

Answer:

$T=2.08s$

Explanation:

From the question we are told that:

Speed of Train $V=3.0m.s$

Angle $\theta=12\textdegree$

Height of window $h_w=0.90m$

Width of window $w_w=2.0m$

The Horizontal distance between B and A from Trigonometric Laws is mathematically given by

$b=\frac{0.9}{tan12}$

$b=4.23$

Therefore

Distance from A-A

$d_a=2.0+4.23$

$d_a=6.23$

Therefore

Time Required to travel trough d is mathematically given as

$T=\frac{d_a}{v}$

$T=\frac{6.23}{3}$

$T=2.08s$

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2 years ago

what is the average gravitational force of attraction between the earth and the sun? the earth averages a distance of about 150

Tanzania [10]

Answer:

B

Explanation:

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2 years ago

Calculate the wave length of a water wave with a speed of 20 m/s and a frequency of 2.5 Hz

12345 [234]

Wavelength of the water wave is 8 m

Explanation:

Wavelength measures the distance between two successive crests or troughs of the wave. It is given by the following equation

λ = v/f, where f is the frequency, v is the velocity of the wave

Here, v = 20 m/s and f = 2.5 Hz

⇒ λ = 20/2.5

= 8 m

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2 years ago

The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th

MaRussiya [10]

Answer:

a). $P=78.4W$

b). $P=392kPa$

c.) It must be at the bottom

Explanation:

Given:

Volume flow $V_f=2.0x10^{-4}m^3/s$

Well depp $h=40.m$

a.

The power output of the pum

$W=F*d$

$F=m*g$

$m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg$

$W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2$

$W=78.4J$

$P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W$

b.

The pressure of difference the pum

Δ $P=p*g*y'$

Δ $P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa$

$P=392kPa$

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way

4 0

3 years ago

In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

Y_Kistochka [10]

Answer: F = 1391 N

Explanation:

The information given to you are:

Mass M = 1300 kg

Acceleration a = 1.07 m/s^2

The magnitude of the force striking the building will be

F = ma

Where

F = force

Substitute mass M and acceleration a into the formula

F = 1300 × 1.07

F = 1391 N

Therefore, the wrecking ball strikes the building with a force of 1391 N

3 0

3 years ago