<h3>Hello there!</h3>
Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.
The equation commonly accepted to find the answer to questions like these is the specific heat formula.
The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.
The information given:
m = 187 grams
c = specific heat capacity of water, or in this case 1 calorie, or 4.184 joules (which is what we will be using)
ΔT = 80 degrees
Now just plug everything in to solve.
Q = 187 * 4.184 * 80
Q = 62592.64
So you have your answer: 62592.64 joules.
Hope this helped!
Answer: The volume is decreasing at a rate of 80 cm3/min
Explanation: Please see the attachments below
Answer:
The answer to your question is: V2 = 1 l
Explanation:
Data
P1 = 200 kPa
P2 = 300 kPa
V1 = 1.5 l
V2 = ?
Formula
P1V1 = P2V2
V2 = (P1V1) / P2
V2 = (200 x 1.5) / 300
V2 = 1 l
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hope this helps !
