How much heat is required to heat 0.1 g of ice at −30 C to steam at 100 C? Use the approximate values below for your calculation.
cice = 2 J/g · C
csteam = 2 J/g · C
cwater = 4 J/g · C
Hvap = 2, 260 J/g
Hfus = 340 J/g
306 J
See this question. Same type.
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Well, let's calculate the heat required step by step, shall we? Let's first bring the ice from -30°C to 0°C:
Heat required to heat ice = mass × specific heat capacity × temperature change
= 0.1 g × 2 J/g · C × (0°C - (-30°C))
= 0.1 g × 2 J/g · C × 30°C
= 0.1 g × 60 J/C
= 6 J
Now, let's calculate the heat required to melt the ice at 0°C:
Heat required to melt ice = mass × heat of fusion
= 0.1 g × 340 J/g
= 34 J
After the ice has melted, we need to heat the water from 0°C to 100°C:
Heat required to heat water = mass × specific heat capacity × temperature change
= 0.1 g × 4 J/g · C × (100°C - 0°C)
= 0.1 g × 4 J/g · C × 100°C
= 40 J
Lastly, we need to convert the water at 100°C to steam at 100°C:
Heat required to vaporize water = mass × heat of vaporization
= 0.1 g × 2260 J/g
= 226 J
Now, let's add up all the heats required in each step:
Total heat required = 6 J + 34 J + 40 J + 226 J
= 306 J
So, approximately 306 J of heat is required to heat 0.1 g of ice at -30°C to steam at 100°C. Good thing this calculation is not part of a stand-up routine, or I'm not sure people would be laughing!
To find the amount of heat required to heat 0.1 g of ice at -30°C to steam at 100°C, we need to consider the different phases and temperature ranges involved.
We can break down the heating process into the following steps:
1. Heating the ice from -30°C to 0°C:
- This step requires the heat to raise the temperature of ice from -30°C to 0°C.
- The specific heat capacity of ice is given as cice = 2 J/g · C.
- The temperature change (∆T) is 0°C - (-30°C) = 30°C.
- The amount of heat required can be calculated using the formula: Q = m * c * ∆T.
Substituting the values:
Q1 = (0.1 g) * (2 J/g · C) * (30°C)
= 6 J
2. Melting the ice at 0°C to water at 0°C:
- This step requires the heat to melt the ice into liquid water.
- The heat of fusion is given as Hfus = 340 J/g.
- The amount of heat required can be calculated using the formula: Q = m * Hfus.
Substituting the values:
Q2 = (0.1 g) * (340 J/g)
= 34 J
3. Heating the water from 0°C to 100°C:
- This step requires the heat to raise the temperature of water from 0°C to 100°C.
- The specific heat capacity of water is given as cwater = 4 J/g · C.
- The temperature change (∆T) is 100°C - 0°C = 100°C.
- The amount of heat required can be calculated using the formula: Q = m * c * ∆T.
Substituting the values:
Q3 = (0.1 g) * (4 J/g · C) * (100°C)
= 40 J
4. Converting water at 100°C to steam at 100°C:
- This step requires the heat to convert water from a liquid state to a gaseous state.
- The heat of vaporization is given as Hvap = 2260 J/g.
- The amount of heat required can be calculated using the formula: Q = m * Hvap.
Substituting the values:
Q4 = (0.1 g) * (2260 J/g)
= 226 J
To find the total amount of heat required, we add up the individual amounts: Qtotal = Q1 + Q2 + Q3 + Q4.
Substituting the values:
Qtotal = 6 J + 34 J + 40 J + 226 J
= 306 J
Therefore, approximately 306 Joules of heat are required to heat 0.1 g of ice at -30°C to steam at 100°C.