how much heat is required to change 400g of ice at -20c into steam at 120c
heat= massice*Heatfusion+masswater*c*100+ masswater*Heatvaporization + masssteam*cs*20
Add the heat required for five separate processes:
(1) Heating the ice from -20 to 0 C
(2) Melting the ice at 0 C
(3) Heating the water to 100 C
(4) Evaporating the water at 100 C
(5) Heating the steam from 100 to 120 C.
For (1) and (5) you will need the specific heats of ice and steam. Look them up.
For (2), use the heat of fusion of 80 calories per gram. Multiply by the 400 g. You require 32,000 calories
For (5), use the heat of vaporization of vaporizing the liquid water, 540 cal/g
For (3), the heat required is 400*1*100 = 40,000 calories.
To determine the amount of heat required to change 400g of ice at -20°C into steam at 120°C, we need to consider several different steps:
Step 1: Heating the ice from -20°C to 0°C:
The specific heat capacity of ice is 2.09 J/g°C, and the temperature change is 0°C - (-20°C) = 20°C. Therefore, the heat required for this step is:
Q1 = mass × specific heat capacity × temperature change
= 400g × 2.09 J/g°C × 20°C
= 16,720 J
Step 2: Melting the ice at 0°C:
To change ice into water (at the melting point), we need to calculate the heat of fusion. The heat of fusion of ice is 334 J/g. Thus, the heat required for this step is:
Q2 = mass × heat of fusion
= 400g × 334 J/g
= 133,600 J
Step 3: Heating the water from 0°C to 100°C:
The specific heat capacity of water is 4.18 J/g°C, and the temperature change is 100°C - 0°C = 100°C. Therefore, the heat required for this step is:
Q3 = mass × specific heat capacity × temperature change
= 400g × 4.18 J/g°C × 100°C
= 167,200 J
Step 4: Boiling the water at 100°C:
To change water into steam, we need to calculate the heat of vaporization. The heat of vaporization of water is 2260 J/g. Thus, the heat required for this step is:
Q4 = mass × heat of vaporization
= 400g × 2260 J/g
= 904,000 J
Step 5: Heating the steam from 100°C to 120°C:
The specific heat capacity of steam is 2.03 J/g°C, and the temperature change is 120°C - 100°C = 20°C. Therefore, the heat required for this step is:
Q5 = mass × specific heat capacity × temperature change
= 400g × 2.03 J/g°C × 20°C
= 16,240 J
To find the total heat required, we sum up the values from each step:
Total heat = Q1 + Q2 + Q3 + Q4 + Q5
= 16,720 J + 133,600 J + 167,200 J + 904,000 J + 16,240 J
= 1,237,760 J
Therefore, it would take 1,237,760 J of heat to change 400g of ice at -20°C into steam at 120°C.
To find out how much heat is required to change 400g of ice at -20°C into steam at 120°C, we need to consider the different stages of the phase change.
First, we need to calculate the heat required to raise the temperature of the ice from -20°C to its melting point at 0°C. This can be done using the specific heat capacity of ice.
Q1 = mass × specific heat capacity × change in temperature
= 400g × 2.09 J/g°C × (0°C - (-20°C))
= 400g × 2.09 J/g°C × 20°C
= 16,720 J
Next, we need to calculate the heat required to change the ice at 0°C into water at 0°C, known as the latent heat of fusion.
Q2 = mass × latent heat of fusion
= 400g × 333.5 J/g
= 133,400 J
After that, we need to calculate the heat required to raise the temperature of the water from 0°C to 100°C. We use the specific heat capacity of water for this calculation.
Q3 = mass × specific heat capacity × change in temperature
= 400g × 4.18 J/g°C × (100°C - 0°C)
= 400g × 4.18 J/g°C × 100°C
= 167,200 J
Now, we need to calculate the heat required to change the water at 100°C into steam at 100°C, known as the latent heat of vaporization.
Q4 = mass × latent heat of vaporization
= 400g × 2260 J/g
= 904,000 J
Finally, we need to calculate the heat required to raise the temperature of the steam from 100°C to 120°C. We use the specific heat capacity of steam for this calculation.
Q5 = mass × specific heat capacity × change in temperature
= 400g × 2.03 J/g°C × (120°C - 100°C)
= 400g × 2.03 J/g°C × 20°C
= 16,240 J
To find the total heat required, we add up all the individual heat values:
Total heat = Q1 + Q2 + Q3 + Q4 + Q5
= 16,720 J + 133,400 J + 167,200 J + 904,000 J + 16,240 J
= 1,237,560 J
Therefore, it would take approximately 1,237,560 Joules of heat to change 400g of ice at -20°C into steam at 120°C.