How much heat would it take to change 1.0kg of water at 20°C to steam at 110°C.
I can't get the right answer that is 2.6×10^6J. Please tell me how can I solve this problem, I want to learn how to solve These problem plz help me.
you have three parts:
a. Heat the water to 100C
b. vaporize the water at 100C
c. heat the steam to 110C
a. heat=m*cwater*(100-20)
b. heat=m*Hv
c. heat=m*csteam*(110-100)
cwater is specific heat capacity of water
Hv is the heat of vaporization of water.
Csteam is specific heat capacity of steam (it is not the same as water)
For (a)336000J
(B) 2255j
(C) 20000j
I added them and the answer is 358255j
Which is totally different from the correct answer ....am I doing it right ?
No.
You have to use units here. What units of mass, specific heat did you use? Notice the units in the table for specific heat, and heat vaporization. That is where You erred.
OK I converted °C to K but the answer now is 4304530J
Did you solved the problem? Is it 2.6×10^6J?
I don't know where I am doing it wrong...I checked my work thrice but still the answer is wrong.
You did not convert celcius to kelvins. They have the same unit.
I will do
a) for you
heat=1kg*cw*(80C)
in my table, cw is is in J/gram*C
heat=1000g*4.186J/gC*80C== 334880J
b) heat=mass*Hv
my table has Hv as 2260 J /g, again mass units is grams
Heat=1000*2260=2,260,000 J
See my point? you have to include units, Then, when you get the three answers, THEY HAVE to be in the same units to add. Yours was NOT.
But my table have specific heat in J/kg.k
And latent heat in kJ/kg
That's why I only converted °c to k and
kJ into J
Can you please tell me what value of specific heat did you used for q3?
To solve this problem, you can use the concept of heat transfer and the specific heat capacities of water in different phases. Here is the step-by-step process to solve the problem:
1. Determine the heat required to raise the temperature of water from 20°C to 100°C.
To calculate the heat transfer, we use the formula: Q = mcΔT
where Q is the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we have water with a mass of 1.0kg, a specific heat capacity of 4.18 J/g·°C, and a temperature change of 100°C - 20°C = 80°C.
Converting the mass from kg to grams:
mass = 1.0kg × 1000g/kg = 1000g
Calculating the heat transfer:
Q1 = 1000g × 4.18 J/g·°C × 80°C = 334,400 J
So, it requires 334,400 Joules of heat to raise the temperature of 1.0kg of water from 20°C to 100°C.
2. Determine the heat required to convert water at 100°C to steam at 100°C.
To transition from water at 100°C to steam at 100°C, we need to consider the heat of vaporization. The heat of vaporization for water is approximately 2.26 × 10^6 J/kg.
So, multiplying the heat of vaporization by the mass of water:
Q2 = 1.0kg × 2.26 × 10^6 J/kg = 2.26 × 10^6 J
3. Determine the heat required to raise the temperature of steam from 100°C to 110°C.
Using the formula Q = mcΔT, we have:
Q3 = 1.0kg × 2.0 J/g·°C × 10°C = 20,000 J
4. Calculate the total heat required by summing up the individual heat transfers:
Total heat = Q1 + Q2 + Q3
Total heat = 334,400 J + 2.26 × 10^6 J + 20,000 J
Total heat = 2.6 × 10^6 J
Therefore, it takes 2.6 × 10^6 Joules of heat to change 1.0kg of water at 20°C to steam at 110°C.
Make sure to verify your values and units along the way to ensure accuracy.