How much heat is lost when 12.4 g of H2O g at 110 degrees C are converted to liquid at 50.0 degrees C? (The specific heat of steam is 2.02 J/g degrees C.

Could someone write the steps?
Equation?

To calculate the amount of heat lost, we can use the equation:

q = mcΔT

Where:
q is the amount of heat lost or gained
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

Given:
Mass of H2O (m): 12.4 g
Specific heat of steam (c): 2.02 J/g·°C
Initial temperature of steam (T1): 110 °C
Final temperature of liquid (T2): 50 °C

Now, let's calculate the amount of heat lost step-by-step:

Step 1: Calculate the heat lost when H2O is cooled from 110 °C to 0 °C:
q1 = mcΔT
q1 = (12.4 g)(2.02 J/g·°C)(110 °C - 0 °C)
q1 = 2664.08 J

Step 2: Calculate the heat lost when H2O freezes from 0 °C to -100 °C:
During this step, the temperature remains constant, so ΔT = 0.
q2 = mcΔT
q2 = (12.4 g)(2.02 J/g·°C)(0 °C - (-100 °C))
q2 = 0 J

Step 3: Calculate the heat lost when H2O is cooled from -100 °C to -50 °C:
q3 = mcΔT
q3 = (12.4 g)(2.02 J/g·°C)(-50 °C - (-100 °C))
q3 = 1248.8 J

The total heat lost is the sum of the heat lost in each step:
q_total = q1 + q2 + q3
q_total = 2664.08 J + 0 J + 1248.8 J
q_total = 3912.88 J

Therefore, the amount of heat lost when 12.4 g of H2O at 110 °C is converted to liquid at 50.0 °C is 3912.88 J.

To calculate the heat lost when 12.4 g of H2O gas at 110 degrees C is converted to liquid at 50.0 degrees C, you need to use the equation:

Q = m * C * ΔT

where:
Q is the amount of heat lost or gained
m is the mass of the substance
C is the specific heat capacity of the substance
ΔT is the change in temperature

In this case, the mass of water (H2O) is 12.4 g, the specific heat capacity of water vapor (steam) is 2.02 J/g degrees C, and the temperature change is from 110 degrees C to 50.0 degrees C.

Here are the steps to calculate the heat lost:

1. Calculate the heat lost when the water vapor cools down from 110 degrees C to 100 degrees C:
Q1 = m * C * ΔT1
Q1 = 12.4 g * 2.02 J/g degrees C * (100 degrees C - 110 degrees C)

2. Calculate the heat lost when the water condenses from water vapor to liquid at 100 degrees C:
Q2 = m * C * ΔT2
Q2 = 12.4 g * 4.18 J/g degrees C * (100 degrees C - 0 degrees C)

3. Calculate the heat lost when the liquid water cools down from 100 degrees C to 50.0 degrees C:
Q3 = m * C * ΔT3
Q3 = 12.4 g * 4.18 J/g degrees C * (50.0 degrees C - 0 degrees C)

4. Add up the three heat values to get the total heat lost:
Q = Q1 + Q2 + Q3

Remember to keep track of units and convert if necessary.