What mass of liquid water a t room temperature (25°C) can be raised to its boiling point with the addition of 40.0 kJ of energy?
to get the heat released or absorbed,
Q = mc(T2-T1)
where
m = mass of substance (units in g)
c = specific heat capacity (units in J/g-K)
T2 = final temperature
T1 = initial temperature
*note: units of temp doesn't matter if it's in degree Celsius or Kelvin, as long as T1 and T2 have the same units (since it's difference in temperature or delta,T)
**note: if Q is (-), heat is released and if (+), heat is absorbed
for water, c = 4.184 J/g-K
substituting,
40 * 10^3 = m*(4.184)*(100 - 25)
40000/(4.184)(75) = m
m = 127.47 g
hope this helps~ :)
Jai why did you multiply 40 by 10^3
Converting it to joules
Since it is 40 kilojoules; the ratio for kilojoules to joules is 1 to 1000, in other words by multiplying 40 by 10^3 (.0001), it is the same as dividing by 1000 to convert the kilojoules to joules.
Well, let's think about this. Heating water requires a certain amount of energy to raise its temperature. Once you add enough energy to reach the boiling point, the water starts to change from a liquid to a gas.
Now, to calculate the mass of water that can be raised to its boiling point with 40.0 kJ of energy, we need to use the specific heat capacity of water. The specific heat capacity of water is about 4.18 J/g°C.
Since we want to raise the temperature from 25°C to the boiling point, which is about 100°C, the change in temperature is 100°C - 25°C = 75°C.
Using the formula Q = m × c × ΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can solve for the mass.
40,000 J = m × 4.18 J/g°C × 75°C
m = 40,000 J / (4.18 J/g°C × 75°C)
After doing the math, we find that the mass of water that can be raised to its boiling point with 40.0 kJ of energy is approximately 191 grams.
So, you would need about 191 grams of water to go from room temperature to its boiling point with 40.0 kJ of energy. Just don't forget to bring a towel!
To determine the mass of liquid water that can be raised to its boiling point with the addition of 40.0 kJ of energy, we need to use the specific heat capacity and the heat of vaporization of water.
First, let's calculate the amount of energy required to raise the temperature of the water from room temperature (25°C) to its boiling point (100°C).
The specific heat capacity of water is 4.18 J/g°C, which means it takes 4.18 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
The temperature change required is:
ΔT = boiling point temperature - initial temperature
ΔT = 100°C - 25°C
ΔT = 75°C
The energy required to raise the temperature of the water is obtained by multiplying the specific heat capacity by the mass of water and the temperature change.
q1 = (mass of water) × (specific heat capacity) × ΔT
Since the specific heat capacity is given in J/g°C, we need to convert the energy required to raise the temperature to joules. 1 kJ = 1000 J.
So, q1 = (mass of water) × (specific heat capacity) × ΔT / (1000 J/kJ)
The energy required to vaporize the water (boiling point to steam) is 40.0 kJ.
Knowing that the heat of vaporization of water is 2260 J/g, we use the equation:
q2 = (mass of water) × (heat of vaporization)
The total energy required is:
Total energy = q1 + q2
Now, we can substitute the given values into the equations to calculate the mass of water:
40.0 kJ = (mass of water) × (specific heat capacity) × ΔT / (1000 J/kJ) + (mass of water) × (heat of vaporization)
To solve for the mass of water, we can rearrange the equation:
mass of water = 40.0 kJ / [(specific heat capacity) × ΔT / (1000 J/kJ) + (heat of vaporization)]
Plugging in the values:
mass of water = 40.0 kJ / [(4.18 J/g°C) × 75°C / (1000 J/kJ) + 2260 J/g]
Calculating the mass of water will give us the answer to the question.