Calculate the amount of heat required to completely convert 1kg of water at 0°C to the boiling point at 100°C?
Give me answer
To calculate the heat required to convert water from 0°C to 100°C, we need to consider two processes:
1. Heating the water from 0°C to 100°C at constant pressure:
The specific heat capacity of water is 4.186 J/g°C. Since we have 1kg of water, the heat required to raise its temperature by 1°C is:
Q1 = mass * specific heat capacity * temperature change
= 1000g * 4.186 J/g°C * (100°C - 0°C)
= 418,600 J
2. Phase change from liquid water at 100°C to steam at 100°C:
The specific heat of vaporization of water is 2,260 J/g. Therefore, the heat required to convert the water at its boiling point into steam at the same temperature is:
Q2 = mass * specific heat of vaporization
= 1000g * 2,260 J/g
= 2,260,000 J
The total heat required is the sum of Q1 and Q2:
Total heat required = Q1 + Q2
= 418,600 J + 2,260,000 J
= 2,678,600 J
Therefore, the amount of heat required to completely convert 1kg of water from 0°C to 100°C is 2,678,600 Joules (J).
To calculate the amount of heat required to completely convert water from 0°C to its boiling point at 100°C, we need to consider two steps:
1. Heating the water from 0°C to 100°C while it remains in the liquid phase.
2. Converting the water from a liquid state at 100°C to a gaseous state (steam) at 100°C.
Step 1: Heating the water from 0°C to 100°C.
To calculate the heat required to raise the temperature of 1kg of water from 0°C to 100°C, we use the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius (J/g°C).
First, convert the mass of water from kilograms to grams:
1kg = 1000 grams.
Next, calculate the heat required to raise the temperature of water:
Heat = mass × specific heat capacity × temperature change
Heat = 1000g × 4.18 J/g°C × (100°C - 0°C)
Step 2: Converting the water to steam at 100°C.
To calculate the heat required to convert 1kg of water at its boiling point from a liquid to a gas (steam), we use the heat of vaporization for water. The heat of vaporization of water is approximately 2,260 Joules per gram (J/g).
Calculate the heat required to convert 1kg of water to steam:
Heat = mass × heat of vaporization
Heat = 1000g × 2260 J/g
Now, we can add the heat from both steps to get the total heat required:
Total heat = Heat from step 1 + Heat from step 2
Substituting the values:
Total heat = (1000g × 4.18 J/g°C × (100°C - 0°C)) + (1000g × 2260 J/g)
After evaluating this equation, you will find the total amount of heat required to completely convert 1kg of water from 0°C to the boiling point at 100°C.
Q = mc change T
have to look up specific heat of water...