What is the total amount of heart absorbed by 100.0 grams of water when the temperature of the water is increased from 30.0 degrees Celcius to 45.0 degrees Celcius?

To calculate the amount of heat absorbed by water, you need to use the formula:

Q = m * C * ΔT

Where:
Q = Heat absorbed (in joules)
m = Mass of the water (in grams)
C = Specific heat capacity of water (4.18 J/g°C)
ΔT = Change in temperature (in °C)

Given:
m = 100.0 grams
C = 4.18 J/g°C
ΔT = 45.0°C - 30.0°C = 15.0°C

Let's substitute the values into the formula:

Q = 100.0 g * 4.18 J/g°C * 15.0°C

Q = 6270 J

Therefore, the total amount of heat absorbed by 100.0 grams of water is 6270 joules.

To find the total amount of heat absorbed by water when its temperature is increased, you can use the formula:

Q = m * C * ΔT

where:
Q is the heat absorbed,
m is the mass of the water (in grams),
C is the specific heat capacity of water (approximately 4.18 J/g°C),
ΔT is the change in temperature (final temperature - initial temperature).

In this case, the mass of the water is given as 100.0 grams, the specific heat capacity of water is approximately 4.18 J/g°C, and the change in temperature is 45.0°C - 30.0°C = 15.0°C.

Now we can calculate the heat absorbed by substituting the values into the formula:

Q = 100.0 g * 4.18 J/g°C * 15.0°C
Q = 6270 J

Therefore, the total amount of heat absorbed by 100.0 grams of water when the temperature is increased from 30.0°C to 45.0°C is 6270 joules (J).

q = mass water x specific heat water x (Tfinal-Tinitial)