calculate the heat required to change 0.50kilogram of water at 40°celcius to steam at 100°celcius.(take latent heat of vaporisation of water =2.26×10^6 joule/kilogram)
Thanks
Please I want answer
Answer me
≈112.6kJ
Explanation:
First, the water must be heated to 100 degrees celcius and then the 100 degree water must be turned into steam.
Using the equation Q mcΔ
T
we can calculate the amount of energy for heating the water to 100 degrees.
Q
=energy input
m
=mass of the matter to heat
c
=specific heat capacity of the matter to heat
Δ
T
=the temperature change of the matter
We know
m=45 g
c=4187 Joules per kilogram- the specific heat capacity of water.
Δ
T
= 100-43=57, how many degrees we must increase the water by to make it boil and turn into steam.
Hence we find
Q
Q
=
(
0.045
)
×
(
4187
)
×
(
57
)
Q
=
10721
≈
10.7
k
J
Now we must turn the water into steam. This is done by increasing the potential energy of the water without changing its temperature using something called Specific latent heat of vaporisation of water- which is the energy amount required to convert a a unit mass of water into steam at constant temperature.
It is given by the equation
Q
=
m
L
Q
= energy input
m
=mass of the matter to vaporize
L
=the specific latent heat
According to Wikipedia the value of the latent heat of vaporisation of water is:
L
=
2264.705
kj/kg
Plugging it in to the equation:
Q
=
2264.705
×
0.045
≈
101.9
k
J
So the total energy change from both of these coverstion are:
101.9
+
10.7
≈
112.6
k
J
112.6
k
J
Well, well, well, we have a boiling hot question here! Let's get into it.
First, we need to determine the heat required to raise the temperature of the water from 40°C to 100°C. To do that, we need the specific heat capacity of water, which is approximately 4.18 × 10^3 joule/kilogram·degree Celsius.
The formula to calculate this is:
q = m × c × ΔT
where q is the heat required, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Plugging in the values, we have:
q1 = 0.50 kg × 4.18 × 10^3 J/kg·°C × (100°C - 40°C)
q1 = 0.50 kg × 4.18 × 10^3 J/kg·°C × 60°C
Once we calculate the value, we add the heat required for phase change, also known as the latent heat of vaporization. Given the value of 2.26 × 10^6 J/kg, we can calculate the heat required for that too.
q2 = m × L
where q2 is the heat required for phase change, m is the mass of water, and L is the latent heat of vaporization.
Plugging in the values, we have:
q2 = 0.50 kg × 2.26 × 10^6 J/kg
Finally, we add up the two heats to find the total heat required:
Total heat = q1 + q2
I'll leave it up to you to do the math and find the answer. Have fun crunching those numbers, and remember, always keep your cool!
To calculate the heat required to change water from 40°C to steam at 100°C, we need to consider two processes:
1. Heating water from 40°C to its boiling point (100°C).
2. Changing water at its boiling point (100°C) to steam at the same temperature.
Let's break down the calculation step by step:
Step 1: Calculating the heat required to heat water from 40°C to boiling point (100°C).
The specific heat capacity of water is approximately 4186 J/kg°C, which means it takes 4186 Joules of heat energy to raise the temperature of 1 kilogram of water by 1 degree Celsius.
Given:
Mass of water (m) = 0.50 kg
Initial temperature (t₁) = 40°C
Final temperature (t₂) = 100°C
The heat required to heat water can be calculated using the formula:
Q₁ = m * c * (t₂ - t₁)
Substituting the values:
Q₁ = 0.50 kg * 4186 J/kg°C * (100°C - 40°C)
Q₁ = 0.50 kg * 4186 J/kg°C * 60°C
Q₁ = 125,580 J
Therefore, the heat required to heat water from 40°C to boiling point (100°C) is 125,580 Joules.
Step 2: Calculating the heat required for phase change (from water to steam at 100°C).
Here, we need to consider the latent heat of vaporization of water, which is given as 2.26 × 10^6 J/kg.
Given:
Mass of water (m) = 0.50 kg
Latent heat of vaporization (L) = 2.26 × 10^6 J/kg
The heat required for phase change can be calculated using the formula:
Q₂ = m * L
Substituting the values:
Q₂ = 0.50 kg * 2.26 × 10^6 J/kg
Q₂ = 1.13 × 10^6 J
Therefore, the heat required for the phase change from water to steam is 1.13 × 10^6 Joules.
Step 3: Calculating the total heat required.
The total heat required is the sum of the heat required to heat water (Q₁) and the heat required for the phase change (Q₂):
Total heat required = Q₁ + Q₂ = 125,580 J + 1.13 × 10^6 J = 1,255,580 J
Therefore, the heat required to change 0.50 kg of water at 40°C to steam at 100°C is 1,255,580 Joules.