how many joules of energy must be removed from 325g water at 35 degress celsius to form ice at 0 degress celsius?
q1 to remove T from 35 to zero.
q1 = mass H2O x specific heat x (0-35)
q2 = heat to remove energy to freeze the water.
q2 = mas H2O x heat fusion.
Total Q = q1 + q2.
Well, to answer your question, we need to calculate the energy required to cool the water from 35 degrees Celsius to 0 degrees Celsius, and then the energy required to freeze it. But don't worry, I'm here to make it a little more fun!
First, we need to cool down the water from 35 degrees Celsius to 0 degrees Celsius. That's like asking water to take a chill pill! So, considering the specific heat capacity of water is 4.186 J/g·°C, we multiply it by the mass of the water (325g) and by the temperature change (35°C - 0°C) to find how much energy is required to cool it down.
Next, we need to freeze the water, or in other words, make it turn into ice. This is a cool transformation, if I may say so! The heat of fusion of water is 334 J/g, and since we have 325g of water, we multiply the two together to find the energy required to freeze it.
Now, let's sum it up! So, to cool down the water from 35°C to 0°C and then freeze it, you would need to remove a total of... drumroll, please... joules of energy. But hey, that's just the scientific side of it. Now let's see if we can find a way to make ice with a snap of our fingers!
To calculate the amount of energy that needs to be removed from the water to form ice, we can use the formula:
Q = m * ΔT * C
where:
Q is the amount of energy
m is the mass of the water
ΔT is the change in temperature
C is the specific heat capacity of water
First, let's calculate ΔT:
ΔT = final temperature - initial temperature
ΔT = 0°C - 35°C
ΔT = -35°C
The specific heat capacity of water (C) is 4.18 J/g°C.
Now, we can substitute the values into the formula:
Q = 325g * -35°C * 4.18 J/g°C
Q = -48875 J
Therefore, approximately 48875 Joules of energy must be removed from the water to form ice at 0 degrees Celsius.
To answer this question, we need to calculate the amount of energy required to cool down the water from 35 degrees Celsius to 0 degrees Celsius, which is the phase transition point between liquid water and ice.
The equation we'll use is:
q = m * ΔT * c
Where:
q is the amount of energy (in joules)
m is the mass of the water (in grams)
ΔT is the change in temperature (in degrees Celsius)
c is the specific heat capacity of water
1. First, let's calculate the energy required to cool down the water from 35 degrees Celsius to 0 degrees Celsius.
ΔT1 = 0 °C - 35 °C = -35 °C
2. Next, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g·°C.
3. Now, we can plug the values into the equation:
q1 = m * ΔT1 * c
q1 = 325 g * (-35 °C) * 4.18 J/g·°C
Calculating this equation will give you the result in joules of energy required to cool down the water from 35 degrees Celsius to 0 degrees Celsius.