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vlabodo [156]
2 years ago
11

8.0 Kg of water is heater from 20°C to 40°C. The specific heat capacity of water is 4200 J/Kg °C. What is the energy supplied to

the water?
Physics
1 answer:
koban [17]2 years ago
6 0

Answer:

672000 J.

Explanation:

From the question,

q = cm(t₂-t₁)............................ Equation 1

Where q = heat energy supplied to the water, c = specific heat capacity of water, m = mass of water, t₂ = final temperature, t₁ = Initial temperature.

Given: m = 8.0 kg, c = 4200 J/kg°C, t₂ = 40°C, t₁ = 20°C

Substitute these values into equation 1

q = 8×4200×(40-20)

q = 8×4200×20

q = 672000 J.

Hence the energy suplied to the water is 672000 J.

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Credit: Department of Energy

5 0
2 years ago
What's at the bottom of the black hole? Explain .​
Brilliant_brown [7]

Answer:

Unknown

Explanation:

By definition, we can't observe what's inside there, because no light – no information of any kind – can escape a black hole. But astrophysical theories suggest that, at the core of a black hole, all the black hole's mass is concentrated into a tiny point of infinite density. This point is known as a singularity.

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2 years ago
You need to make a very dilute solution of sodium phosphate dihydrate (MW 142 g/mol, FW 178 g/mol). How many grams of sodium pho
dlinn [17]

Answer:

first question is 178 g of sodium phosphate dihydrate. second question is 0.25 mL

Explanation:

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6 0
2 years ago
An oven uses 4500 watts of power in 2 minutes, how much work was done?
jolli1 [7]
First of all we state the formula
Power=work done/time
we can rearrange this formula as well
work done=power x time

Since the SI unit of time is in seconds we change the minutes to seconds
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Using our formula (work done=power x time) we simply put in the values
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4 0
2 years ago
The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use
saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

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Where

N = Near point

l = distance between the object lens and eye lens

f_0= Focal length

f_e= Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

M = \frac{25*10}{3*5}

M = 16.67\approx 17\\

Therefore the correct answer is C.

3 0
3 years ago
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