how much heat is released wen 10g of steam at 100 degrees c is cooled to 10g of liquid water at 100degrees c?

q = mass steam x heat vaporization = ?

To calculate the heat released when steam is cooled to water, you need to use the specific heat equation. The specific heat equation is given by:

Q = mcΔT

Where:
Q is the heat released or absorbed (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in Joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).

In this case, the mass (m) is given as 10 grams and the change in temperature (ΔT) is from 100 degrees Celsius (steam) to 100 degrees Celsius (liquid water), which means there is no change in temperature. Therefore, ΔT = 0.

Now, we need to find the specific heat capacity (c) for water and steam. The specific heat capacity of steam and liquid water is different. The specific heat capacity of steam is 2.03 J/g°C, and the specific heat capacity of liquid water is 4.18 J/g°C.

Let's calculate the heat released by steam when it is cooled to liquid water at the same temperature:

For steam:
Q1 = (mass of steam) x (specific heat capacity of steam) x (change in temperature)
Q1 = 10 g x 2.03 J/g°C x (100 - 100) °C
Q1 = 0 J

As ΔT is 0, the heat released will be zero.

Hence, no heat is released when 10g of steam at 100 degrees Celsius is cooled to 10g of liquid water at 100 degrees Celsius.