how much energy is released when a 5600 gram sample of water cools from 99 degress celsius to 28 degrees celcius
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To calculate the amount of energy released when the water cools from 99 degrees Celsius to 28 degrees Celsius, we need to use the equation:
Q = mcΔT
Where:
Q = heat energy (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = temperature change (in °C)
First, we need to calculate the temperature change (ΔT):
ΔT = (final temperature - initial temperature)
= (28°C - 99°C)
= -71°C
Since the temperature change is negative, it means that heat is being released.
Next, we can substitute the values into the equation to find the energy released:
Q = (5600 g) * (4.18 J/g°C) * (-71°C)
= -16685120 J
Therefore, the amount of energy released when a 5600 gram sample of water cools from 99 degrees Celsius to 28 degrees Celsius is approximately -16,685,120 Joules. Note that the negative sign indicates that heat is being released.
To calculate the amount of energy released when a sample of water cools, you can use the formula:
Q = mcΔT
Where:
Q = Energy released or absorbed (in joules)
m = Mass of the substance (in grams)
c = Specific heat capacity of the substance (in J/g°C)
ΔT = Change in temperature (in °C)
In this case, we need to find the energy released when a 5600 gram sample of water cools from 99 degrees Celsius to 28 degrees Celsius. The specific heat capacity of water is approximately 4.18 J/g°C.
Let's substitute the given values into the formula:
Q = (5600 g) x (4.18 J/g°C) x (99°C - 28°C)
Q = 5600 g x 4.18 J/g°C x 71°C
Now, calculate the result:
Q = 1666464 J
Therefore, approximately 1,666,464 Joules of energy are released when a 5600 gram sample of water cools from 99 degrees Celsius to 28 degrees Celsius.
Note the correct spelling of celsius.
q = mass x specific heat x delta T.
q = 5600 g x 4.184 J/g*C x (99-28)