How much heat (in BTU) would be required to heat and vaporize 100 lb of water from 25°C to saturated steam at 1 atm? Assume heat capacity is constant at 1 BTU/lb-°F and the latent heat of vaporization is 1,000 BTU/lb
To calculate the total heat required to heat and vaporize 100 lb of water from 25°C to saturated steam at 1 atm, we need to consider the following steps:
1. Heating the water from 25°C to its boiling point at 1 atm.
2. Vaporizing the water at its boiling point.
Let's break down the calculations step-by-step:
Step 1: Heating the water from 25°C to its boiling point at 1 atm.
To calculate the heat required for this step, we'll use the formula:
Heat = mass * specific heat * change in temperature
Given:
- Mass of water = 100 lb
- Specific heat of water = 1 BTU/lb-°F
- Initial temperature = 25°C
- Final temperature = boiling point of water at 1 atm = 100°C (assuming the boiling point is rounded to 100°C at 1 atm)
First, we need to convert the initial and final temperatures to Fahrenheit:
Initial temperature = (25°C × 9/5) + 32°F = 77°F
Final temperature = (100°C × 9/5) + 32°F = 212°F
Using the formula, we can calculate the heat required to heat the water:
Heat = 100 lb * 1 BTU/lb-°F * (212°F - 77°F) = 13,500 BTU
Step 2: Vaporizing the water at its boiling point.
To calculate the heat required to vaporize the water, we'll use the formula:
Heat = mass * latent heat of vaporization
Given:
- Mass of water = 100 lb
- Latent heat of vaporization = 1,000 BTU/lb
Using the formula, we calculate the heat required to vaporize the water:
Heat = 100 lb * 1,000 BTU/lb = 100,000 BTU
Finally, to find the total heat required, we sum up the heat from both steps:
Total heat required = Heat from Step 1 + Heat from Step 2
Total heat required = 13,500 BTU + 100,000 BTU = 113,500 BTU
Therefore, it would require 113,500 BTU of heat to heat and vaporize 100 lb of water from 25°C to saturated steam at 1 atm.
To calculate the amount of heat required to heat and vaporize water, there are two steps involved: heating the water to its boiling point and then vaporizing it.
Step 1: Heating the water to its boiling point.
To heat the water from 25°C to its boiling point, you need to calculate the heat required to raise the temperature by that amount.
The specific heat capacity of water is given as 1 BTU/lb-°F, which means it requires 1 BTU of heat energy to raise the temperature of 1 pound of water by 1 degree Fahrenheit.
1 pound of water is equivalent to 100 BTU per 1°F temperature rise.
In this case, we need to raise the temperature from 25°C to the boiling point, which is 100°C. To convert from Celsius to Fahrenheit, you can use the formula: °F = °C * (9/5) + 32.
So, the temperature rise in Fahrenheit is (100°C * 9/5) + 32 - (25°C * 9/5) + 32 = 180°F.
To find the heat required, we multiply the temperature rise by the weight of the water:
Heat required for heating = 100 lb * 180°F * 1 BTU/lb-°F = 18,000 BTU.
Step 2: Vaporizing the water.
To calculate the heat required to vaporize the water at its boiling point, the latent heat of vaporization comes into play. The latent heat of vaporization is given as 1,000 BTU/lb.
So, the heat required to vaporize the water is:
Heat required for vaporization = 100 lb * 1,000 BTU/lb = 100,000 BTU.
Finally, to find the total heat required to heat and vaporize the water, we sum up the heat required for heating and vaporization:
Total heat required = Heat required for heating + Heat required for vaporization
= 18,000 BTU + 100,000 BTU
= 118,000 BTU.
Therefore, 118,000 BTU of heat would be required to heat and vaporize 100 lb of water from 25°C to saturated steam at 1 atm.