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At 1 atm, how much energy is required to heat 49.0 g of H2O(s) at –10.0 °C to H2O(g) at 129.0 °C?

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3 Answers

Hi Diane,


When you heat ice at −10.0 °C until it becomes superheated steam at 129.0 °C, you are proceeding through several steps, and you have to take the time and make the calculations to account for each of those steps. In a general sense, adding thermal energy to H2O or any other pure substance can do one of two things. First, it can raise the temperature of the substance. In that case, the heat exchanged is called sensible heat. Second, it can cause a phase transition in the substance. If this happens, the heat exchanged is called latent heat. For your question, you need to add together the contributions of all of these steps for 49.0 g of H2O:

step 1) Find the sensible heat needed to increase the temperature of H2O(s) (ice) by 10.0 °C from −10.0 °C to 0.0 °C.

step 2) Find the latent heat needed to melt ice at 0 °C into liquid water at 0 °C. This is called the latent heat of fusion.

step 3) Find the sensible heat needed to increase the temperature of H2O(l) (liquid water) by 100.0 °C from 0.0 °C to 100.0 °C.

step 4) Find the latent heat needed to boil water at 100 °C into steam at 100 °C. This is called the latent heat of vaporization.

step 5) Find the sensible heat needed to increase the temperature of H2O(g) (water vapor) by 29.0 °C from 100.0 °C to 129.0 °C.


The three steps involving sensible heat (#1, #3, and #5) all use the formula:

Q = cp · m · ΔT

but solid ice, liquid water, and gaseous water vapor each have their own specific heat, so you'll need three different values of cp. Fortunately, these values are easy to locate for H2O. Also be sure to use the correct change in temperature that corresponds to that particular step in the process.

 

The two steps involving latent heat (#2 and #4)  both use the formula:

Qm · L

The latent heat of fusion is different from the latent heat of vaporization, so you'll need two different values of L. Again, these values for the latent heat of fusion and vaporization are easy to find for water.

 

Add the contributions together for each substep, and you'll have the energy required for the entire process: Qtotal = Q1 + Q2 + Q3 + Q4 + Q5.

Comments

Marc's right: there are five steps to this problem, all with a different constant you need to look up. Beautifully put, Marc.

Add up the amounts of energy spent (heating up ice) + (melting ice into liquid water) + (raising the temperature of liquid water)  + (boiling water into steam) + (increasing the temperature of the steam). The sum will be the total energy invested.

A common mistake is to forget that water travels through three different phases, so we need five steps. Another common mistake is to forget that ice and steam have different specific heats from liquid water.

Comment

Recall that Q=c m dT, where Q is the energy required to raise a mass m dT degrees an c is the specific heat capacity of the substance. Furthermore recall that the specific heat capacity of water is 1cal/g/C.

Thus we have Q = 1cal/g/C * 49.0g * (129.0C-(-10C)) = 1cal/g/C * 49.0g * 139.0C

Using the commutative and associative properties of multiplication we get

Q=1*49.0*139.0 * cal/g/C*g*C

the g and C cancel out leaving us with Q=1*49.0*139.0 * cal or

Q=6811cal

 

The amount of energy need to heat a subject can be expressed by the following formula:

     Q = cp · m · ΔT  , 

   where       Q = amount of heat

                   cp = specific heat

                   m = mass

                   ΔT = change in temperature (T2 - T1)

You are given the following information for heating water (H2O):

     mass of water = m = 49.0 g

     change in temperature = ΔT = T2 - T1 = 129 - (-10) = 129 + 10 = 139 °C

Recall that the specific heat of water is:     1.0 cal/g·°C

Using the heat formula above, we arrive at the following:

     Q = (1.0 cal/g·°C)(49.0 g)(139 °C)

        = 6,811 calories

Since it is known that there are 4.184 Joules per calorie, we can convert the above answer to Joules as follows:

       (6,811 cals)·(4.184 J/cal) = 28,497.224 J

                                            ≈ 28,497 J

There is 1 kiloJoule per 1,000 Joules. With this we can convert the above answer into kiloJoules as follows:

       (28,497 J)(1 kJ/1,000 J) = 28.497 kJ

                                          ≈ 28.5 kJ

Thus, 28.5 kJ of energy is required to heat 49.0 g of H2O from -10.0 °C to 129.0 °C.