How much heat is required to convert 3kg of ice at -5 degees C into steam at 125 degrees C?
The answer is 9.6x10^6 J but I don't know how to get there. Thank you!
heat ice to 0ºC
melt ice
heat water to 100ºC
boil water
heat steam to 125ºC
specific heats - ice, water, steam
... heats of fusion and vaporization
To calculate the amount of heat required to convert a substance from one state to another, we need to consider the following steps:
1. Heating the ice from -5 degrees C to its melting point at 0 degrees C.
2. Melting the ice at 0 degrees C into water at 0 degrees C.
3. Heating the water from 0 degrees C to its boiling point at 100 degrees C.
4. Converting the water at 100 degrees C into steam at 100 degrees C.
5. Heating the steam from 100 degrees C to the desired temperature of 125 degrees C.
Let's calculate the heat required for each step:
1. Heating the ice from -5 degrees C to 0 degrees C:
The specific heat capacity of ice is 2.09 J/g·°C.
The heat required can be calculated using the formula: Q = m × c × ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q1 = 3 kg × 2090 J/kg·°C × (0 - (-5)) = 31350 J
2. Melting the ice at 0 degrees C:
The heat required to melt ice is the product of its heat of fusion and the mass of ice.
The heat of fusion for ice is 334,000 J/kg.
Q2 = 3 kg × 334000 J/kg = 1002000 J
3. Heating the water from 0 degrees C to 100 degrees C:
The specific heat capacity of water is 4.18 J/g·°C.
Q3 = 3 kg × 4180 J/kg·°C × (100 - 0) = 1254000 J
4. Converting water at 100 degrees C into steam at 100 degrees C:
The heat required to vaporize water is the product of its heat of vaporization and the mass of water.
The heat of vaporization for water is 2,260,000 J/kg.
Q4 = 3 kg × 2260000 J/kg = 6780000 J
5. Heating the steam from 100 degrees C to 125 degrees C:
The specific heat capacity of steam is 2.03 J/g·°C.
Q5 = 3 kg × 2030 J/kg·°C × (125 - 100) = 91450 J
Now, let's calculate the total heat required:
Total Heat = Q1 + Q2 + Q3 + Q4 + Q5
Total Heat = 31350 J + 1002000 J + 1254000 J + 6780000 J + 91450 J
Total Heat = 9,253,800 J
Therefore, approximately 9,253,800 J of heat is required to convert 3 kg of ice at -5 degrees C into steam at 125 degrees C.
To calculate the heat required to convert a substance from one phase to another, you need to consider the following steps:
1. Determine the amount of heat required to raise the temperature of ice from -5 degrees Celsius to its melting point (0 degrees Celsius). This can be calculated using the specific heat capacity of ice, which is 2.09 J/g°C.
Heat = mass x specific heat capacity x temperature change
The mass of ice is given as 3 kg, but we need to convert it to grams:
Mass of ice = 3 kg x 1000 g/kg = 3000 g
Temperature change = 0°C - -5°C = 5°C
Heat = 3000 g x 2.09 J/g°C x 5°C
2. Determine the amount of heat required to convert the ice at 0 degrees Celsius to water at the same temperature. This is known as the heat of fusion for water, which is 334 J/g.
Heat = mass x heat of fusion
Heat = 3000 g x 334 J/g
3. Determine the amount of heat required to raise the temperature of the water from 0 degrees Celsius to its boiling point (100 degrees Celsius).
Heat = mass x specific heat capacity x temperature change
The mass of water is equal to the mass of ice (after melting) because there is no change in mass during phase change.
Heat = 3000 g x 4.18 J/g°C x 100°C
4. Determine the amount of heat required to convert the water at 100 degrees Celsius to steam at the same temperature. This is known as the heat of vaporization for water, which is 2260 J/g.
Heat = mass x heat of vaporization
Heat = 3000 g x 2260 J/g
Now, add up the heat values calculated in each step to find the total heat required:
Total heat = Heat step 1 + Heat step 2 + Heat step 3 + Heat step 4
Note: Make sure you use consistent units (grams) throughout the calculations.
When you perform these calculations, you will get the total amount of heat required to convert 3 kg of ice at -5 degrees Celsius into steam at 125 degrees Celsius.