If 682000 kJ could be transferred with 100% efficiency to water, what volume of water, in liters, could be heated from 8.6 degrees Celsius to 60.0 degrees Celsius as a result?
To determine the volume of water that can be heated, we need to use the specific heat formula:
Q = mcΔT
where:
Q is the heat energy transferred (in joules)
m is the mass of the substance (in kilograms)
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
First, convert the given energy from kJ to J by multiplying by 1000:
682000 kJ * 1000 J/kJ = 682000000 J
Next, we can calculate the mass of the water. The specific heat capacity of water is approximately 4.186 J/g°C or 4186 J/kg°C:
m = Q / (c * ΔT)
m = 682000000 J / (4186 J/kg°C * (60.0°C - 8.6°C))
m = 682000000 J / (4186 J/kg°C * 51.4°C)
m ≈ 3240.05 kg
Finally, convert the mass of water from kilograms to liters. Since the density of water is approximately 1 g/mL or 1000 kg/m³:
V = m / ρ
V = 3240.05 kg / 1000 kg/m³
V ≈ 3.24005 m³
V ≈ 3240.05 L
Therefore, the volume of water that can be heated from 8.6°C to 60.0°C with 100% efficiency is approximately 3240.05 liters.