Calculate the heat released from 61.5 g of steam at 111 degrees C is converted to water at 39 degrees C. Assume the specific heat of water is 4.184 J/g*C, the specific heat of steam is 1.99 J/g*C and the enthalpy.H.vap= 40.79 KJ/mol of water

To calculate the heat released, we need to consider the two steps involved in the process: condensing the steam to water at the same temperature and then cooling the water to the final temperature.

Step 1: Converting steam to water at the same temperature (111 °C)
The heat released during this step can be calculated using the formula:

q1 = m * ΔH
where:
q1 is the heat released (in joules)
m is the mass of the steam (in grams)
ΔH is the enthalpy of vaporization (in joules/gram)

First, we need to convert the mass of steam from grams to moles:
moles of steam = mass of steam / molar mass of water

The molar mass of water is approximately 18.015 g/mol.

moles of steam = 61.5 g / 18.015 g/mol

Next, we convert moles of steam to joules:
q1 = moles of steam * ΔH

The enthalpy of vaporization is given in KJ/mol, so we need to convert it to J/g:
ΔH = 40.79 KJ/mol * (1000 J/KJ) / (moles of steam)

Finally, we can calculate q1:
q1 = moles of steam * ΔH

Now, we have the heat released during the phase change from steam to water at the same temperature.

Step 2: Cooling the water to the final temperature (39 °C)
The heat released during this step can be calculated using the formula:

q2 = m * c * ΔT
where:
q2 is the heat released (in joules)
m is the mass of water (in grams)
c is the specific heat of water (in joules/gram*C)
ΔT is the change in temperature (final temperature - initial temperature)

In this case, the initial temperature is 111 °C and the final temperature is 39 °C.

Now, we can calculate q2:

q2 = 61.5 g * 4.184 J/g*C * (39°C - 111°C)

Finally, we add q1 and q2 to obtain the total heat released:

Total heat released = q1 + q2

Now, you can calculate the answer using the above steps and formulas.