I'm not sure but it should be 7.31 times 2 times 3.14 or pi
7.31 x 2 x 3.14 = 45.9068
If you need to round to the nearest tenth it will be 45.9
If you need to round to the nearest hundredth it will be 45.91
Answer:
Classifying stars according to their spectrum is a very powerful way to begin to understand how they work. As we said last time, the spectral sequence O, B, A, F, G, K, M is a temperature sequence, with the hottest stars being of type O (surface temperatures 30,000-40,000 K), and the coolest stars being of type M (surface temperatures around 3,000 K). Because hot stars are blue, and cool stars are red, the temperature sequence is also a color sequence. It is sometimes helpful, though, to classify objects according to two different properties. Let's say we try to classify stars according to their apparent brightness, also. We could make a plot with color on one axis, and apparent brightness on the other axis, like this:
Explanation:
Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
