How long would it take a 20Kw heater to warm up 1000 litres of water from 15 degrees to 55 degrees in hours? specific heat of water = 4.18kJ.
also
How much heat would a block of ice melting at 200Kg per hour absorb?
thanks if anyone can help!
How much heat is needed?
q = mass H2O x specific heat H2O x delta T.
Then J/s x s = 20,000 watt. Solve for s and convert to hours.
To calculate the time it would take for a 20 kW heater to warm up 1000 liters of water from 15 degrees to 55 degrees in hours, you can use the equation:
Q = m * c * ΔT
where:
Q is the heat energy required in joules,
m is the mass of water in kilograms,
c is the specific heat capacity of water in kJ/kg°C, and
ΔT is the temperature change in degrees Celsius.
First, convert the volume of water from liters to kilograms:
m = 1000 liters * 1 kg/L = 1000 kg
Next, determine the temperature change:
ΔT = 55°C - 15°C = 40°C
Plugging these values into the equation:
Q = (1000 kg) * (4.18 kJ/kg°C) * (40°C) = 167,200 kJ
Since the heater has a power output of 20 kW, you can convert the energy in kilojoules to the time it takes to produce that amount of energy:
Time = Energy / Power
Time = (167,200 kJ) / (20 kW) = 8.36 hours
Therefore, it would take approximately 8.36 hours for the 20 kW heater to warm up 1000 liters of water from 15 degrees to 55 degrees.
As for the second question, to determine how much heat a block of ice melting at a rate of 200 kg per hour would absorb, you can use the equation:
Q = m * ΔHf
where:
Q is the heat energy in joules,
m is the mass of the ice in kilograms, and
ΔHf is the latent heat of fusion for water, which is the energy required to change the state from solid to liquid, typically given as kJ/kg.
Assuming the ΔHf for ice is 334 kJ/kg, then:
Q = (200 kg) * (334 kJ/kg) = 66,800 kJ
Therefore, the block of ice melting at a rate of 200 kg per hour would absorb approximately 66,800 kJ of heat energy.