calculate the number of joules released when 175.0 grams of water are cooled from 100.0 to 27.5 celcius
Well, crashing temperature is a cool thing to talk about! To calculate the energy released, we can use the heat capacity of water and the equation:
Q = m * C * ΔT
Where:
Q is the energy released
m is the mass of water (175.0 grams)
C is the specific heat capacity of water
ΔT is the change in temperature (100.0 - 27.5 Celsius)
The specific heat capacity of water is approximately 4.18 J/g°C.
Now, let's do some math to find out the joules released when water cools down...
Q = 175.0 g * 4.18 J/g°C * (100.0 - 27.5) °C
Q = 175.0 g * 4.18 J/g°C * 72.5 °C
Let's plug that into the calculator and calculate:
Q ≈ 51,378.75 J
So, when 175.0 grams of water are cooled from 100.0 to 27.5 Celsius, approximately 51,378.75 joules are released. But don't worry, those joules are just playing a cooling role!
To calculate the number of joules released when cooling water, you'll need to use the specific heat capacity and the formula:
Q = m * C * ΔT
Where:
Q is the amount of heat released (in joules),
m is the mass of the substance (in grams),
C is the specific heat capacity of the substance (in J/g·°C), and
ΔT is the change in temperature (in °C).
First, let's find the change in temperature.
ΔT = Tfinal - Tinitial
ΔT = 27.5°C - 100.0°C
ΔT = -72.5°C
Now, let's find the specific heat capacity of water.
The specific heat capacity of water is approximately 4.184 J/g·°C.
Now, we can substitute the values into the formula:
Q = m * C * ΔT
Q = 175.0 g * 4.184 J/g·°C * -72.5°C
Q ≈ -53748 Joules
Therefore, approximately 53,748 Joules of heat are released when 175.0 grams of water are cooled from 100.0 to 27.5 degrees Celsius.
To calculate the amount of energy released when cooling water, you need to use the specific heat capacity of water and its mass.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
First, calculate the amount of energy required to decrease the temperature from 100.0°C to 0°C:
Mass of water = 175.0 grams
Temperature change = (100°C - 0°C) = 100°C
Energy change = mass × specific heat capacity × temperature change
Energy change = 175.0 g × 4.18 J/g°C × 100°C
Energy change = 73150 J
Next, calculate the amount of energy required for the water to cool down further from 0°C to 27.5°C:
Mass of water = 175.0 grams
Temperature change = (27.5°C - 0°C) = 27.5°C
Energy change = mass × specific heat capacity × temperature change
Energy change = 175.0 g × 4.18 J/g°C × 27.5°C
Energy change = 20143.75 J
To find the total energy released when cooling the water from 100.0°C to 27.5°C, you need to add the energy changes together:
Total energy released = Energy change from 100°C to 0°C + Energy change from 0°C to 27.5°C
Total energy released = 73150 J + 20143.75 J
Total energy released = 93293.75 J
Therefore, the total energy released when cooling 175.0 grams of water from 100.0°C to 27.5°C is approximately 93293.75 Joules.