How much heat in kilojoules is required to warm 10.0 g of ice, initially at -10.0 ∘C, to steam at 115 ∘C. The heat capacity of ice is 2.09 J/g∘C and that of steam is 1.84 J/g∘C.
2.83 kJ
To calculate the total heat required to warm the ice to steam, we need to consider three steps:
1. Heating the ice from -10.0 °C to 0 °C
2. Melting the ice at 0 °C
3. Heating the liquid water from 0 °C to 115 °C
Step 1: Heating the ice from -10.0 °C to 0 °C
The heat required for this step can be calculated using the formula:
q = m × C × ΔT
where q is the heat required, m is the mass of the ice, C is the heat capacity of ice, and ΔT is the change in temperature.
q1 = 10.0 g × 2.09 J/g.°C × (0 °C - (-10.0 °C))
q1 = 10.0 g × 2.09 J/g.°C × 10.0 °C
q1 = 209 Joules
Step 2: Melting the ice at 0 °C
The heat required to melt the ice can be calculated using the formula:
q = m × ΔHf
where q is the heat required, m is the mass of the ice, and ΔHf is the heat of fusion of ice (334 J/g).
q2 = 10.0 g × 334 J/g
q2 = 3340 Joules
Step 3: Heating the liquid water from 0 °C to 115 °C
The heat required for this step can be calculated using the formula:
q = m × C × ΔT
where q is the heat required, m is the mass of the water, C is the heat capacity of steam, and ΔT is the change in temperature.
q3 = 10.0 g × 1.84 J/g.°C × (115 °C - 0 °C)
q3 = 10.0 g × 1.84 J/g.°C × 115 °C
q3 = 2116 Joules
Now, we can sum up the heat required for each step to find the total heat:
Total heat = q1 + q2 + q3
Total heat = 209 Joules + 3340 Joules + 2116 Joules
Total heat = 5665 Joules
To convert the total heat to kilojoules:
Total heat in kilojoules = Total heat in joules / 1000
Total heat in kilojoules = 5665 Joules / 1000
Total heat in kilojoules ≈ 5.67 kilojoules
Therefore, approximately 5.67 kilojoules of heat is required to warm 10.0 g of ice, initially at -10.0 °C, to steam at 115 °C.
To find the amount of heat required to warm the ice to steam, we need to consider each step of the phase change separately.
Step 1: Heating the ice to its melting point (0°C)
The heat required can be calculated using the formula:
Q = m * c * ΔT
where:
Q = heat energy
m = mass of the substance
c = heat capacity
ΔT = change in temperature
Given:
mass (m) = 10.0 g
heat capacity (c) of ice = 2.09 J/g∘C
change in temperature (ΔT) = (0°C - (-10.0°C)) = 10.0°C
Plugging in the values:
Q1 = 10.0 g * 2.09 J/g∘C * 10.0°C
Q1 = 209 J
So, 209 J of heat energy is required to warm the ice to its melting point.
Step 2: Melting the ice to water at 0°C
To melt the ice, we need to calculate the heat energy using the formula:
Q = m * ΔHf
where:
Q = heat energy
m = mass of the substance
ΔHf = heat of fusion (energy required to change a substance from solid to liquid)
The heat of fusion for ice is 333.55 J/g.
Q2 = 10.0 g * 333.55 J/g
Q2 = 3335.5 J
So, 3335.5 J of heat energy is required to melt the ice.
Step 3: Heating the water from 0°C to 100°C
Using the same formula as in Step 1:
Q3 = m * c * ΔT
Given:
mass (m) = 10.0 g
heat capacity (c) of water = 4.184 J/g∘C (Typical value for liquid water)
change in temperature (ΔT) = (100°C - 0°C) = 100°C
Q3 = 10.0 g * 4.184 J/g∘C * 100.0°C
Q3 = 4184 J
So, 4184 J of heat energy is required to heat the water from 0°C to 100°C.
Step 4: Vaporizing the water to steam at 100°C
To vaporize the water, we need to calculate the heat energy using the formula:
Q = m * ΔHv
where:
Q = heat energy
m = mass of the substance
ΔHv = heat of vaporization (energy required to change a substance from liquid to gas)
The heat of vaporization for water is 2260 J/g.
Q4 = 10.0 g * 2260 J/g
Q4 = 22600 J
So, 22600 J of heat energy is required to vaporize the water.
Step 5: Heating the steam from 100°C to 115°C
Again, using the formula from Step 1:
Q5 = m * c * ΔT
Given:
mass (m) = 10.0 g
heat capacity (c) of steam = 1.84 J/g∘C
change in temperature (ΔT) = (115°C - 100°C) = 15°C
Q5 = 10.0 g * 1.84 J/g∘C * 15.0°C
Q5 = 276 J
So, 276 J of heat energy is required to heat the steam from 100°C to 115°C.
Now, to find the total heat energy required, we add up the individual amounts of heat energy:
Total heat energy = Q1 + Q2 + Q3 + Q4 + Q5
Total heat energy = 209 J + 3335.5 J + 4184 J + 22600 J + 276 J
Total heat energy = 30304.5 J
Converting the answer to kilojoules:
Total heat energy in kilojoules = 30304.5 J / 1000
Total heat energy in kilojoules = 30.3045 kJ
Therefore, approximately 30.3045 kilojoules of heat energy are required to warm 10.0 g of ice, initially at -10.0°C, to steam at 115°C.
heat ice to 0ºC ... 10.0 g *10.0 ºC * 2.09 J/gºC
melt ice ... 10.0 g * 334 J/g
heat water to 100ºC ... 10.0 g * 4.184 J/gºC * 100.0 ºC
boil water ... 10.0 g * 2.23 kJ/g
heat steam to 115ºC ... 10.0 g * 1.84 J/gºC * 15.0 ºC
add the steps