Determine the Final state and temperature of 100g of water originally at 25 degree C after 50 kJ of heat have been added to it.

To determine the final state and temperature of the 100g of water after adding 50 kJ of heat, we need to use the specific heat capacity of water and the equation:

Q = m * c * ΔT

where:
Q is the heat energy (in Joules),
m is the mass of the water (in grams),
c is the specific heat capacity of water (4.18 J/g°C),
ΔT is the change in temperature (in °C).

First, let's convert the heat energy from kJ to Joules:
50 kJ = 50,000 J

Next, let's rearrange the equation to solve for ΔT:
ΔT = Q / (m * c)

ΔT = 50,000 J / (100g * 4.18 J/g°C)

ΔT ≈ 119.61 °C

Now, let's calculate the final temperature:
Final temperature = initial temperature + ΔT
Final temperature = 25 °C + 119.61 °C
Final temperature ≈ 144.61 °C

Therefore, the final state of the water after adding 50 kJ of heat is in its liquid phase, and the final temperature is approximately 144.61 °C.

To determine the final state and temperature of 100g of water after adding 50 kJ of heat, we need to consider the specific heat capacity of water and apply the equation:

q = mcΔT

Where:
q is the amount of heat added (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)

First, let's find the heat capacity of water (c):

The specific heat capacity of water is approximately 4.18 J/g°C.

Now, let's calculate ΔT using the given information:

q = mcΔT

Rearranging the equation to solve for ΔT:

ΔT = q / (mc)

Plugging in the values:
m = 100g
c = 4.18 J/g°C
q = 50 kJ = 50,000 J

ΔT = 50,000 J / (100g * 4.18 J/g°C)

ΔT = 119.14 °C

Now, we can determine the final temperature. Since the water was initially at 25°C, we add the change in temperature (ΔT) to the initial temperature to get the final temperature:

Final temperature = Initial temperature + ΔT

Final temperature = 25°C + 119.14°C

Final temperature = 144.14°C

Therefore, the final state of the 100g of water, originally at 25°C, after adding 50 kJ of heat, is a temperature of 144.14°C.

45

heat needed to raise T from 25C to 100 C.

q = mass H2O x specific heat H2O x (75)
q = approx 32,000 J (BUT THAT'S APPROXIMATE--YOU NEED TO GET A BETTER ANSWER)
That leaves 50,000-32,000 to be used to convert to steam.

q to convert to steam = mass x heat vap
APPROX 18,000 = mass H2O x heat vaporization.
Look up heat vap and solve for mass H2O that can be converted to steam.
You will have part liquid H2O @ 100 C and part steam at 100 C.

The remaining heat can heat the steam to an elevated temperature but no more liquid water can be converted to steam.
Therefore, your answer will be x grams liquid H2O at 100C, y grams steam at some T above 100.