if the temperature of a 500.0g sample of liquid water is raised to 2.00c how much heat is absorbed by the water? The specific heat of liquid water is 4.184J/(g*c)
q = mass water x specific heat water x (Tfinal - Tinitial)
How much energy must be absorbed by 20.0 g of water to increase its temperature from 43.0 °C to 83.0 °C? (C = 4.184 J/g°C)
To determine the amount of heat absorbed by the water, you can use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat absorbed (in joules, J)
m is the mass of the water (in grams, g)
c is the specific heat of water (in joules per gram per degree Celsius, J/(g*C))
ΔT is the change in temperature (in degrees Celsius, °C)
Given:
m = 500.0 g
c = 4.184 J/(g*C)
ΔT = 2.00°C
Substituting the values into the formula:
Q = 500.0 g * 4.184 J/(g*C) * 2.00°C
Calculating the result:
Q = 4184 J * 2.00°C
Q = 8368 J
Therefore, the water absorbs 8368 joules of heat.