How much heat is needed to cause the following irreversible reaction?

200 g of liquid H2O at 50°C is heated to form 200 g of steam at 120°C

q1 = heat needed to raise T from 50 C to 100 C.

q1 = mass H2O x specific heat liquid H2O x (delta T)-----delta T = Tfinal-Tinitial = 100-50 = 50

q2 = heat needed to vaporize liquid H2O at 100 C to vapor(steam) at 100 C.
q2 = mass H2O x heat vaporization.

q3 = heat needed to heat steam at 100 to steam a 120 C.
q3 = mass steam x specific heat steam x delta T----120-100 = 20

Total q needed = q1 + q2 + q3.

To calculate the heat needed for this reaction, we need to consider the specific heat capacity and enthalpy of phase changes for water.

1. Heating water from 50°C to its boiling point:
The specific heat capacity of liquid water is 4.18 J/g°C.
The temperature change is: 100°C (boiling point - initial temperature).
The heat needed for this step is calculated using the formula: q = m * c * ΔT

q = (200 g) * (4.18 J/g°C) * (100°C)
q = 83600 J

2. Vaporizing liquid water at its boiling point:
The enthalpy of vaporization for water is 40.7 kJ/mol.
Since the molar mass of water is 18 g/mol, the enthalpy of vaporization can be expressed as 40.7 kJ/18 g.

The heat needed for this step is calculated using the formula: q = m * ΔHvap

q = (200 g) * (40.7 kJ/18 g)
q = 45222.2 J

Total heat needed = 83600 J + 45222.2 J
Total heat needed = 128,822.2 J

Therefore, approximately 128,822.2 J of heat is needed to cause the given irreversible reaction.

To determine the amount of heat needed for this irreversible reaction, you need to consider the processes involved. The heat required can be calculated by adding the heat required to raise the temperature of water from 50°C to its boiling point and the heat required to convert it to steam at its boiling point.

The specific heat capacity of water is approximately 4.18 J/g°C, which means that 4.18 Joules of heat energy is required to raise the temperature of 1 gram of water by 1 degree Celsius.

First, calculate the heat required to raise the temperature of 200 g of water from 50°C to its boiling point:

Q1 = mass × specific heat capacity × change in temperature
= 200 g × 4.18 J/g°C × (100°C - 50°C)
= 200 g × 4.18 J/g°C × 50°C
= 41800 J

Next, calculate the heat required to convert the water at its boiling point to steam:

The heat of vaporization of water is approximately 2260 J/g. This means that 2260 Joules of heat energy is needed to convert 1 gram of water at its boiling point to steam, without any change in temperature.

Q2 = mass × heat of vaporization
= 200 g × 2260 J/g
= 452000 J

Finally, add the two values to get the total heat required:

Total heat = Q1 + Q2
= 41800 J + 452000 J
= 493800 J

Therefore, approximately 493800 Joules (or 493.8 kJ) of heat energy is needed to cause the irreversible reaction described.