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.
To solve this problem, we can use the formula:
Q = mcΔT
Where:
Q = energy transferred (in Joules)
m = mass (in kilograms)
c = specific heat capacity (in J/kg°C)
ΔT = change in temperature (in °C)
First, we need to calculate the energy transferred (Q) from cooling the water to its freezing point:
Q = (2844 MJ/h) * (40 min / 60 min/h) * (3600 s/h) * (10^6 J/MJ)
= 7.5864 * 10^9 J
Next, we need to calculate the energy required to cool the water from 87°C to 0°C. The specific heat capacity of water is approximately 4.18 J/g°C.
Q = mcΔT
7.5864 * 10^9 J = (m)(4.18 J/g°C)(87°C)
We need to convert the mass to kilograms and the specific heat capacity to J/kg°C:
(7.5864 * 10^9 J) / (4.18 J/g°C) = m * (87°C)
1.818 * 10^9 g = m * (87°C)
Now we can calculate the mass of the water:
m = 1.818 * 10^9 g / 87°C
m ≈ 20.92 * 10^6 g
Finally, we need to convert the mass to kilograms:
m ≈ 20.92 kg
Therefore, the mass of the water/ice is approximately 20.92 kilograms.