Answer the following questions about energy changes associated with heating and cooling a substance. Use the quantities in the table for your calculations.
(table)
Specific heat of water /4.184J/g*K
dealta H vap for water /40.7 kJ/mol
dealta H fus for water /6.02 kJ/mol
dealta H f for CO2 (g) /-393.5 kJ/mol
dealta H f for H20 (l) /-285.8 kJ/mol
dealta H f for CH4 (g) /-74.9 kJ/mol
a. A puddle of water on a sidewalk is heated by the Sun by about 3ºC. The mass of the water is about 2 kg. Estimate the heat absorbed by the water. Assume no water evaporates, and no heat is transferred from the water to either the air or the sidewalk.
q=m*c*DeltaT
q=2000g*4.184J/g*K*276K=2.310 * 10^6 J
This was my answer however 2.310 * 10^6 J my teacher told me I made an error in my calculations.
I see the error. The specific heat of water is Joules per grams CELSIUS (not Kelvin).
q = (2000 g)(4.184 J/gºC)(3ºC) = 25104 J
herp: A celsius degree is exactly the same as a Kelvin. The specific heat is exactly the same per degree C, or per degree K. And finally, the temperature change from freezing water to boiling water is 100C, and 100K. Goodness. Please don't try to help in Chemistry. Thanks.
I also made an error in this part of the question
b. A 2.75 kg block of ice is heated until it completely melts. It is then heated again until it completely changes to water vapor. Is more energy required to melt the ice or to vaporize the liquid water? Show your calculations.
q1(to melt ice) = mass ice x heat fusion.
q2(to raise T to 100 C) = mass H2O x specific heat H2O x (Tfinal-Tinitial).
q2* (to torn liquid H2O at 100 C to steam at 100 C) = mass H2O x heat vaporization. q2 + q2*
=raising temperature from zero to 100 and then turning it to steam.
q1=16.56 kgkJ/mol.
q2=1150.6 J
q2*=111.93 kgkJ/mol.
As it can be seen in my calculations, it takes a lot more energy to vaporize the liquid water than to melt the ice.
Herp, you are totally wrong.
YOu do not see the difference between the absolute scale and changes.
on the absolute scale
a temp of -273K is the same as 0C
but a change temperature from 0K to 273K, is the same change as -273C to OC. A change of 10 C is the same as 10 K.
q3=mass*specificeheat*100=2.75kg*4.19kJ/kgC*100= you do it.But it is not your answer, by 1000
I don't understand what you are saying. I put 2.75kg*4.19kJ/kgC*100 into a calculator and got 1150.6
To determine the heat absorbed by the water, you correctly used the formula q = m * c * ΔT, where q represents the heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
In this case, the mass of the water is 2 kg, and the change in temperature is 3ºC. However, the specific heat capacity of water is given in joules per gram per degree Celsius (J/g°C), so you need to convert the mass of the water from grams to kilograms.
1 kg = 1000 grams
Thus, 2 kg = 2000 grams.
Now, let's calculate the heat absorbed by the water:
q = m * c * ΔT
= 2000 g * 4.184 J/g°C * 3°C
= 25,104 J
Therefore, the correct answer should be 25,104 J, which is equivalent to 2.5104 × 10^4 J.
But 1 C = 274.15 K
If you had shown your work instead of answers we could have seen at a glance what the problem was. As it is you must check your work against mine below. I recognize this as one of the answers I gave you a couple of days ago. You have made errors in q1, q2 and q2*. I didn't check anything else. Also I didn't worry about the significant figures; you must do that.
q1 = mass ice x heat fusion
q1 = 2.75 kg x (1000g/kg) x (1 mol/18g) x 6.02 kJ/mol = 919.7 kJ.
Note: I converted kg ice to mols since you were given heat fusion in kJ/mol
q2 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = 2.75 kg x (1000 g/kg) x 4.184 J/g*C x 100 = 1.1506E6 J = 1.1506E3 kJ = 1150.6 kJ
I used g for mass since specific heat is given in J/g*C
q2* = mass H2O x heat vap
q2* = 2.75 kg x (1000g/kg) x (1 mol/18g) x 40.7 kJ/mol =6218 kJ.
I converted 2.75 kg to g and to mols since heat vap is given to you in kJ/mol.
Note: I didn't check the significant figures. You must do that.
yes, that is true. However, your caclulations leave something to be desired.
q1=mass*Hf=2.75kg*334kJ/kg
I don't know why you figured it per mole (18g of ice), you figured it wrong. And your units kgKJ/mole what is that?
q3=mass*specificeheat*100=2.75kg*4.19kJ/kgC*100= you do it.But it is not your answer, by 1000
Q3=mass*Hv=2.75kg*2260kJ/kg you do it, but is it not your answer.
In short, I don't think you know what you are doing. There are plenty of websites that can help you, google heat fusion, or heat vaporizaiton, or specific heat.