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PilotLPTM [1.2K]
3 years ago
8

How much heat must be absorbed by 375 grams of water to raise its temperature by 25°c

Physics
2 answers:
Nesterboy [21]3 years ago
7 0
<span>the formula q = 375 g * 25 C * 4.186 J / (g*C) = 39,243.75 J q represents the heat in Joules , m the mass in grams, difference of temperature in Celsius degree, and 4.186 J/(g*C) is the specific heat of water( I assume the water is in liquid from and will remain liquid). Approximately 39.24 kJ once you round and transform to kJ..1 kJ=1000J</span>
N76 [4]3 years ago
4 0

Answer:

Explanation:

From H= mc×temp rise

m= mass of water = 375g or 0.375kg

c= heat capacity of water = 4200JKg-1

H= 0.375×4200×25=39375J or 39.375KJ

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An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and
svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

7 0
3 years ago
If one of two interacting charges is doubled, the force between the charges will _____________.
malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

F=\frac{k q1 q2}{r^{2}}

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get F'=\frac{k (2q1) q2}{r^{2}}

F'= 2 F

Thus the force gets doubled.

4 0
3 years ago
Which of the following is a derived unit?<br><br> ampere<br> newton<br> second<br> kilogram
Soloha48 [4]

Newton is your answer.

ampere, second, kilograms, are all base units.

hope this helps

5 0
2 years ago
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