How many grams of water would require 4000 joules of heat to raise its temperature from 65°C to 100°C? The specific heat of water is 4.181?

27.33 g water

Step-by-step explanation:

For this problem, we will need to use the equation for heat: q=mCΔT. Since we are given heat, change in temperature, and specific heat, all we have to do is plug them in to find grams.

4000J=m(4.181J/g°C)(100-65°C)              [find change in temperature]

4000J=m(4.181J/g°C)(35°C)    [divide both sides by specific heat and change in temperature]

m=27.33 g

p(x)= 8 x³ - 36 x² + 54 x - 27

q(x) = 2 x -3

you can check your mistake and then you can choose the correct option.

dividend =   8 x³ - 36 x² + 54 x - 27

divisor = 2 x -3

if you want to find quotient, proceed as follows

q(x)= 0

2 x - 3=0

x =

now , i have used division method , to find quotient and remainder