. A water heater can has a power of 350 W. It is placed in a container filled with 1 litres of water with a temperature of 40 ºC. The heater is then allow to run for 1 hour. How much water will there be left in the container after 1 hour? (Specific heat capacity of water = 4200 J/kg K and specific latent heat of vaporization = 2260 000 J/kg K)
1 liter of water has a mass of 1.00 kg
Energy added in one hour =
3600 s * 350 J/s = 1.26*10^6 J
Energy required to heat all of the water to 100C (the boiling point) =
1 kg*60 K*4200 J/kgK = = 2.52*10^5 J
Energy available to boil water =
1.26*10^6 - 0.252*10^6 = 0.998*10^6 J
Amount that boils away
= 0.998*10^6 J/2.26*10^6 J/kg
= 0.44 kg
The amount left will be 0.56 kg.
Notye: The units of the latent heat of vaporization should be J/kg, NOT J/kg*K
To solve this problem, you would need to calculate the energy supplied by the water heater and then use it to determine the amount of water that is evaporated.
First, let's calculate the energy supplied by the water heater in 1 hour. The power of the water heater is given as 350 W, and the time is 1 hour:
Energy supplied = Power × Time
= 350 W × 1 hour
= 350 watt-hours
Next, we need to convert this energy to joules since the specific heat capacity and specific latent heat of vaporization are given in joules. To convert watt-hours to joules, we use the conversion factor 1 watt-hour = 3600 joules:
Energy supplied (in joules) = 350 watt-hours × 3600 joules/watt-hour
Now that we have the energy supplied by the water heater in joules, we can use it to calculate the amount of water evaporated.
The energy required to raise the temperature of water is given by the equation:
Energy = Mass × Specific heat capacity × Temperature change
In this case, we assume that the temperature of the water heater remains constant at 100 ºC, the boiling point of water. Therefore, the temperature change is:
Temperature change = 100 ºC - 40 ºC
= 60 ºC
We need to find the mass of water that is evaporated. Let's assume the mass to be 'm' kg.
Energy supplied (in joules) = Specific latent heat of vaporization × Mass
Substituting the values we have:
350 watt-hours × 3600 joules/watt-hour = 2260,000 J/kg × m
Now we can solve for 'm':
m = (350 watt-hours × 3600 joules/watt-hour) / (2260,000 J/kg)
≈ 0.557 kg
Therefore, approximately 0.557 kg or 557 grams of water will evaporate during this 1 hour period.
To calculate the amount of water that will be left in the container after 1 hour, subtract the mass evaporated from the initial mass:
Initial mass - Mass evaporated = Final mass
Final mass = 1 kg - 0.557 kg
≈ 0.443 kg or 443 grams
So, approximately 0.443 kg or 443 grams of water will be left in the container after 1 hour.