How many Joules of thermal energy must be added to 5 kg of water at 20 C to bring it to the boiling point (at 100 C)?
Q =c•m•ΔT=4180•5•80 =1.6•10^6 J.
To calculate the amount of thermal energy required to heat the water from 20°C to 100°C, you can use the specific heat capacity formula:
Q = mcΔT,
where Q is the thermal energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius (J/g°C). To use this value, we need to convert the mass of water from kilograms to grams.
Mass of water = 5 kg = 5000 grams
Now, let's calculate the thermal energy required:
Q = (mass of water) × (specific heat capacity of water) × (change in temperature)
ΔT = 100°C - 20°C = 80°C
Q = (5000 g) × (4.18 J/g°C) × (80°C)
Q ≈ 1,672,000 J
Therefore, approximately 1,672,000 Joules of thermal energy must be added to the 5 kg of water at 20°C to bring it to the boiling point at 100°C.
To find the amount of thermal energy required to raise the temperature of water from 20°C to its boiling point at 100°C, we can use the specific heat capacity and the equation:
Q = mcΔT
Where:
- Q is the amount of thermal energy (in Joules)
- 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 find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius. To convert it to kilograms, divide by 1000:
c = 4.18 J/g/°C / 1000 = 0.00418 J/kg/°C
Now we can substitute the values into the formula:
Q = 5 kg * 0.00418 J/kg/°C * (100°C - 20°C)
Q = 5 kg * 0.00418 J/kg/°C * 80°C
Q = 5 kg * 0.3344 J/°C
Q = 1.672 Joules
Therefore, approximately 1.672 Joules of thermal energy must be added to 5 kg of water at 20°C to raise its temperature to its boiling point at 100°C.