This is my first time taking chemistry and I am VERY VERY confused
On the step involved to find out how many calories to warm 58mL of water from 10 degress C to 100 degress C and vaporize it at 100 degress C.
q1 = heat to warm water from 10 to 100.
q1 = mass x specific heat x (Tfinal-Tinitial) where Tf is 100 and Ti is 10 (or delta T = 90).
q2 = heat to vaporize liquid water at 100 to vapor at 100 C.
q2 = mass water x heat vaporization
Total q = q1 + q2.
Which is faster, a bullet with a velocity of 1100. feet per second, or Superman traveling at 1100. miles per hour? Provide the correct velocity for either Superman or the bullet in scientific notation with the correct number of sig figs. You must use the following conversion factors: Density = m/V; 1 L = 1000 mL; 1 L = 1.057 quarts; 1 gallon = 4 quarts; 1 mile = 5280 feet; 1 foot = 12 inches; 1 inch = 2.54 cm; 1 meter = 100 cm; 1 kilometer = 1000 m; 1 g = 0.03527 oz; 1 lb = 16 oz
To determine the number of calories required to warm 58 mL of water from 10 degrees Celsius to 100 degrees Celsius and vaporize it at 100 degrees Celsius, you need to follow these steps:
Step 1: Calculate the energy required to heat the water from 10 degrees Celsius to 100 degrees Celsius.
To determine the energy required to warm the water, you will use the equation:
Q = m * C * ΔT
Where:
Q = Energy (in calories)
m = Mass of the water (in grams)
C = Specific heat capacity of water (1 cal/g °C)
ΔT = Change in temperature (in °C)
First, convert the volume of water (58 mL) to grams. Since the density of water is 1 g/mL, 58 mL of water is equal to 58 grams.
Next, calculate the change in temperature:
ΔT = Final temperature - Initial temperature
ΔT = 100 °C - 10 °C
ΔT = 90 °C
Now, plug in the values into the equation:
Q = 58 g * 1 cal/g °C * 90 °C
Q = 5220 cal
So, it requires 5220 calories to heat the water to 100 degrees Celsius.
Step 2: Calculate the energy required to vaporize the water at 100 degrees Celsius.
To determine the energy required to vaporize the water, you will use the equation:
Q = m * Hvap
Where:
Q = Energy (in calories)
m = Mass of the water (in grams)
Hvap = Heat of vaporization of water (around 540 cal/g)
Using the mass of the water, which is 58 grams, and the heat of vaporization, which is approximately 540 cal/g, plug in the values into the equation:
Q = 58 g * 540 cal/g
Q = 31320 cal
So, it requires 31320 calories to vaporize the water at 100 degrees Celsius.
Step 3: Add the results from Step 1 and Step 2.
To find the total energy required to warm the water from 10 degrees Celsius to 100 degrees Celsius and then vaporize it, simply add the values calculated in Step 1 and Step 2:
Total energy required = Energy to heat water + Energy to vaporize water
Total energy required = 5220 cal + 31320 cal
Total energy required = 36540 cal
Therefore, it requires 36540 calories to warm 58 mL of water from 10 degrees Celsius to 100 degrees Celsius and vaporize it at 100 degrees Celsius.