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3 years ago

How much energy is need to raise 50 kg of water from 45 c to 80c?

2 answers:

8 0

Based on your problem, what you are looking for is the quantity of heat. To solve for it, you will need this formula:

Q = mc(T2-T1)

Where: Q = Quantity of heat
m = mass of the substance
c = Specific heat
T2 = Final temperature
T1 = Initial temperature

Now the specific heat of water is 4.184 J/(g°C), meaning that is how much energy is required to raise the temperature of 1g of liquid water by 1 degree Celsius.

Since your mass is in kilograms, let us convert that into grams, which will be equal to 50,000 grams. Now we can put our given into the equation:

Q = mc(T2-T1)
= 50,000g x 4.184 J/(g°C) x (80°C - 45°C)
= 50,000 g x 4.184 J/(g°C) x 35°C
= 7,322,000 J or 7,322 kJ or 7.322 MJ

gavmur [86]3 years ago

6 0

Energy = mass ×specific heat capacity×change in temperature'
In this case, the mass of water is 50 kg, the specific heat capacity of water is 4184J/kg
the change in temperature is 80-45= 35 degrees celcius
Therefore, heat energy
= 50 ×35 ×4184=7322000 Joules
But 1kJ =1000 Joules
therefore, the energy needed will be 7322 kJ

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3 years ago

You are traveling at 100 km/hr to make it to work on time. You have to be to work in .25 hr or you will be late. You have 16 km

alexgriva [62]

Answer: Yes

Explanation:

Velocity $V$ is defined as the distance traveled $d$ in a specific time $t$ :

$V=\frac{d}{t}$

If you are traveling at $V=100 km/h$ a distance $d=16 m$ , then the time it will take you to be at work is:

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This means you will make it on time, because this time is less than 0.25 h.

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4 years ago

A sewing machine uses 1.6kWh of electricity in one day. If electricity costs 9p per unit, what is the total cost in pence of usi

atroni [7]

Answer:

$\huge\boxed{\sf 1.6\ units = 14.4p}$

Explanation:

Since,

So,

1.6 kWh = 1.6 units

If,

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$\rule[225]{225}{2}$

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2 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

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3 years ago