how much energy is released when a 5600-g sample of water cools from 99 degrees C to 28 degrees C?
q = heat released = mass x specific heat H2O x delta T.
is the answer 1,663,558.4 J
To calculate the energy released when a substance cools, we can use the equation:
Q = mcΔT
where Q is the energy transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
In this case, we are dealing with water, which has a specific heat capacity of approximately 4.18 J/g°C.
Given:
Mass of water (m) = 5600 g
Change in temperature (ΔT) = (28°C - 99°C) = -71°C
Substituting the values into the equation, we have:
Q = (5600 g) × (4.18 J/g°C) × (-71°C)
To simplify, we'll convert the mass to kilograms:
Mass of water (m) = 5.6 kg
Q = (5.6 kg) × (4.18 J/g°C) × (-71°C)
To calculate the energy transferred, simply multiply the numbers together:
Q = -13336.96 J
Therefore, approximately -13,336.96 J of energy is released when a 5600-g sample of water cools from 99 degrees C to 28 degrees C.