A 150L water heater is set to a cutoff temperature of 49 degrees C and that the incoming water is at 298K, calculate the energy in kJ needed to get this 150L water to it's maximum temperature. Please help!
heat energy=mc deltaTemp
= 150kg*specificheatwater*(49+273-298)
look up specific heat of water, and you have it.
Just looking for an answer
To calculate the energy needed to heat the water to its maximum temperature, we need to use the formula:
Q = mcΔT
Where:
Q is the energy (in joules) needed to heat the water
m is the mass of the water (in kilograms)
c is the specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
First, let's convert the volume of water from liters to kilograms. We'll use the density of water, which is approximately 1 kg/L.
Mass (m) = volume (V) × density (d)
= 150 L × 1 kg/L
= 150 kg
Next, we need to convert the change in temperature from Celsius to Kelvin since the temperature values are given in Kelvin.
Change in temperature (ΔT) = cutoff temperature - incoming water temperature
= 49 °C - 25 °C (298 K - 25 °C is approximately 298 K)
= 24 °C (or 24 K)
Now, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C or 4.18 kJ/kg°C.
Now, we can plug in the given values into the formula to calculate the energy (Q) needed:
Q = mcΔT
= 150 kg × 4.18 kJ/kg°C × 24 °C
≈ 150 × 4.18 × 24 kJ
≈ 15,043.2 kJ
Therefore, the energy needed to heat the 150L of water to its maximum temperature is approximately 15,043.2 kJ.