A quantity of water at 87 degrees C is cooled at a rate of 2844 MJ/h. After 40 minutes, the water has changed to ice, having a temperature of 243K. Calculate the mass of the water/ice.
I know 243K = -30 degrees C
I'm assuming I have to use my sensible and latent heat formulas but I would need heat energy in kJ.
53c
To calculate the mass of the water/ice, we can use the energy formula:
\(Q = mc\Delta T + mL\)
Where:
- \(Q\) is the heat energy in joules
- \(m\) is the mass of the water/ice in grams
- \(c\) is the specific heat capacity of water (4.18 J/g°C)
- \(\Delta T\) is the change in temperature in degrees Celsius
- \(L\) is the latent heat of fusion for water (334 J/g)
First, let's calculate the heat energy absorbed by the water to change its temperature from 87°C to 0°C:
\(Q_1 = mc\Delta T\)
Substituting the given values:
\(\Delta T = 0°C - 87°C = -87°C\)
\(Q_1 = m \times 4.18 \, \text{J/g°C} \times (-87) \, \text{°C}\)
To convert this heat energy to kJ, divide by 1000:
\(Q_1 = \frac{m \times 4.18 \, \text{J/g°C} \times (-87) \, \text{°C}}{1000} \, \text{kJ}\)
Next, let's calculate the heat energy absorbed by the water to change its state from liquid to solid ice:
\(Q_2 = mL\)
Substituting the given values:
\(Q_2 = m \times 334 \, \text{J/g}\)
To convert this heat energy to kJ, divide by 1000:
\(Q_2 = \frac{m \times 334 \, \text{J/g}}{1000} \, \text{kJ}\)
We know the total heat energy absorbed by the water is given by the cooling rate and the time:
\(Q_{\text{total}} = \text{cooling rate} \times \text{time}\)
Substituting the given values:
\(Q_{\text{total}} = 2844 \, \text{MJ/h} \times 40 \, \text{min}\)
Note: Convert the time to hours by dividing by 60 (since there are 60 minutes in an hour) before multiplying.
Now, converting MJ to kJ by multiplying by 1000:
\(Q_{\text{total}} = (2844 \times 1000) \, \text{kJ/h} \times (40/60) \, \text{h}\)
The total heat energy absorbed is equal to the sum of \(Q_1\) and \(Q_2\):
\(Q_{\text{total}} = Q_1 + Q_2\)
Substituting the expressions for \(Q_1\) and \(Q_2\) into the equation:
\((2844 \times 1000) \, \text{kJ/h} \times \left(\frac{40}{60}\right) \, \text{h} = \frac{m \times 4.18 \, \text{J/g°C} \times (-87) \, \text{°C}}{1000} \, \text{kJ} + \frac{m \times 334 \, \text{J/g}}{1000} \, \text{kJ}\)
Now, we can solve for \(m\), the mass of the water/ice.