How much heat must be removed from 75.0 grams of water at 90.0 degrees Celsius to cool it to 23.0 degrees celsius?

To calculate the amount of heat that needs to be removed, you can use the equation:

Q = mcΔT

Where Q is the amount of heat transferred, m is the mass of the substance (in this case, water), c is the specific heat capacity of the substance, and ΔT is the change in temperature.

First, let's find the change in temperature:

ΔT = final temperature - initial temperature
ΔT = 23.0°C - 90.0°C
ΔT = -67.0°C

Now, let's find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.

Now, substitute the values into the equation:

Q = (75.0g)(4.18 J/g°C)(-67.0°C)
Q = -21,086.5 J

Note that the negative sign indicates that heat is being removed from the water. The answer is approximately -21,086.5 Joules.