How much heat is required to warm 225g of ice from -46.8 degree celcius to 0.0 degree celcius, melt the ice, warm the water from 0.0 degree celcius to 100.0 degree celcius, boil water, and heat the steam to 173.0 degree celcius.
Figure the heat in each of those steps, then add them.
I will be happy to critique your work.
To calculate the total heat required, we need to consider the different steps involved:
1. Warming the ice from -46.8°C to 0.0°C:
The specific heat capacity of ice is approximately 2.09 J/g°C. So, the heat required is:
Q1 = mass × specific heat capacity × change in temperature
= 225g × 2.09 J/g°C × (0.0 - (-46.8)°C)
2. Melting the ice:
The heat required to melt the ice is given by the equation:
Q2 = mass × heat of fusion
= 225g × 334 J/g (heat of fusion of ice)
3. Warming the water from 0.0°C to 100.0°C:
The specific heat capacity of water is approximately 4.18 J/g°C. The heat required is:
Q3 = mass × specific heat capacity × change in temperature
= 225g × 4.18 J/g°C × (100.0 - 0.0)°C
4. Boiling the water:
The heat required to boil water is given by:
Q4 = mass × heat of vaporization
= 225g × 2260 J/g (heat of vaporization of water)
5. Heating the steam from 100.0°C to 173.0°C:
The specific heat capacity of steam is approximately 2.03 J/g°C. The heat required is:
Q5 = mass × specific heat capacity × change in temperature
= 225g × 2.03 J/g°C × (173.0 - 100.0)°C
Now, we can calculate the total heat required:
Total Heat = Q1 + Q2 + Q3 + Q4 + Q5
Plug in the values and calculate each step to find the total heat required.
To calculate the total heat required for the entire process, you need to consider each step individually and then sum up the heat for each step.
Step 1: Warming ice from -46.8°C to 0.0°C
To calculate the heat required to warm the ice, you need to use the specific heat capacity of ice, which is 2.09 J/g°C.
The equation to calculate the heat energy is:
Q = m * c * ΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Mass (m) = 225g
Specific heat capacity (c) = 2.09 J/g°C
Change in temperature (ΔT) = 0.0°C - (-46.8°C) = 46.8°C
Plugging in the values into the equation:
Q1 = 225g * 2.09 J/g°C * 46.8°C
Q1 ≈ 21948 J
So, the heat required to warm the ice from -46.8°C to 0.0°C is approximately 21948 Joules.
Step 2: Melting the ice
To calculate the heat required to melt the ice, you need to use the heat of fusion of water, which is 334 J/g.
The equation to calculate the heat energy is:
Q = m * ΔHf
where Q is the heat energy, m is the mass, and ΔHf is the heat of fusion.
Given:
Mass (m) = 225g
Heat of fusion (ΔHf) = 334 J/g
Plugging in the values into the equation:
Q2 = 225g * 334 J/g
Q2 = 75150 J
So, the heat required to melt the ice is 75150 Joules.
Step 3: Warming water from 0.0°C to 100.0°C
To calculate the heat required to warm the water, you need to use the specific heat capacity of water, which is 4.18 J/g°C.
The equation to calculate the heat energy is:
Q = m * c * ΔT
Given:
Mass (m) = 225g
Specific heat capacity (c) = 4.18 J/g°C
Change in temperature (ΔT) = 100.0°C - 0.0°C = 100.0°C
Plugging in the values into the equation:
Q3 = 225g * 4.18 J/g°C * 100.0°C
Q3 = 93750 J
So, the heat required to warm the water from 0.0°C to 100.0°C is 93750 Joules.
Step 4: Boiling water and heating steam to 173.0°C
To calculate the heat required to boil the water and heat it to 173.0°C, you need to use the heat of vaporization of water, which is 2260 J/g.
The equation to calculate the heat energy is:
Q = m * ΔHv
Given:
Mass (m) = 225g
Heat of vaporization (ΔHv) = 2260 J/g
Plugging in the values into the equation:
Q4 = 225g * 2260 J/g
Q4 = 508500 J
So, the heat required to boil the water and heat the steam to 173.0°C is 508500 Joules.
To calculate the total heat required for the entire process, you sum up the heat for each step:
Total heat energy = Q1 + Q2 + Q3 + Q4
Total heat energy = 21948 J + 75150 J + 93750 J + 508500 J
Total heat energy ≈ 700348 J
Therefore, approximately 700348 Joules of heat is required for the entire process.