How much heat (in kJ) is required to warm 12.0g of ice, initially at -12.0C, to steam at 111.0?C The heat capacity of ice is 2.09J/(g*C) and that of steam is 2.01J/(g*C). I got 5.64 but I did something wrong apparently.
Well, it seems like you're in hot water with this question! Let's have some fun figuring it out together.
To warm the ice to 0°C, we need to use the formula:
q = m * c * ΔT
where q is the heat required, m is the mass, c is the heat capacity, and ΔT is the change in temperature.
So, for the first step of warming the ice from -12.0°C to 0°C:
q1 = 12.0 g * 2.09 J/(g°C) * (0°C - (-12.0°C))
q1 = 12.0 g * 2.09 J/(g°C) * 12.0°C
q1 = 299.52 J
Now, for the second step of melting the ice to water at 0°C, we use the heat of fusion:
q2 = m * ΔHf
The heat of fusion for ice is approximately 334 J/g, so:
q2 = 12.0 g * 334 J/g
q2 = 4008 J
Next, we need to warm the water from 0°C to 100°C:
q3 = 12.0 g * 4.18 J/(g°C) * (100°C - 0°C)
q3 = 12.0 g * 4.18 J/(g°C) * 100°C
q3 = 5016 J
And finally, we need to use the heat of vaporization to convert the water at 100°C to steam at 111.0°C:
q4 = m * ΔHv
The heat of vaporization for water is approximately 2260 J/g, so:
q4 = 12.0 g * 2260 J/g
q4 = 27120 J
Adding up all the heat required in each step, we get:
q = q1 + q2 + q3 + q4
q = 299.52 J + 4008 J + 5016 J + 27120 J
q = 36231.52 J
Now, let's convert this to kJ:
q = 36231.52 J * (1 kJ / 1000 J)
q = 36.23 kJ
So, the correct answer is approximately 36.23 kJ. I hope this clears things up and puts a smile on your face!
To calculate the amount of heat required to warm the ice to steam, we need to consider the heating of each phase individually: ice, liquid water, and steam.
Step 1: Heating ice to 0°C (heat required to raise the temperature of ice)
Q1 = mass × specific heat capacity × temperature change
= 12.0 g × 2.09 J/(g·°C) × (0°C - (-12.0°C))
= 12.0 g × 2.09 J/(g·°C) × 12.0°C
= 299.52 J
Step 2: Melting of ice at 0°C (heat required to change ice to liquid water)
Q2 = heat of fusion × mass of ice
= 333.55 J/g × 12.0 g
= 4002.6 J
Step 3: Heating liquid water from 0°C to 100°C
Q3 = mass × specific heat capacity × temperature change
= 12.0 g × 4.18 J/(g·°C) × (100°C - 0°C)
= 12.0 g × 4.18 J/(g·°C) × 100.0°C
= 5016.0 J
Step 4: Vaporization of liquid water at 100°C (heat required to change water to steam)
Q4 = heat of vaporization × mass of water
= 2257 J/g × 12.0 g
= 27084 J
Step 5: Heating steam from 100°C to 111°C
Q5 = mass × specific heat capacity × temperature change
= 12.0 g × 2.01 J/(g·°C) × (111.0°C - 100.0°C)
= 12.0 g × 2.01 J/(g·°C) × 11.0°C
= 532.92 J
Total heat required:
Q_total = Q1 + Q2 + Q3 + Q4 + Q5
= 299.52 J + 4002.6 J + 5016.0 J + 27084 J + 532.92 J
= 36644.04 J
Converting J to kJ:
Q_total = 36.64404 kJ
Therefore, the amount of heat required to warm 12.0 g of ice to steam is 36.644 kJ, not 5.64 kJ as you calculated.
To calculate the amount of heat required to warm a substance from one temperature to another, you need to follow the steps of the heat transfer process and use the equation:
Q = m * c * ΔT
Where:
Q is the heat transferred (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (final temperature - initial temperature) in °C
Let's break down the steps to calculate the heat required to warm the ice and then convert it to steam.
1. Heat required to warm the ice to its melting point (0°C):
Q1 = m * c1 * ΔT1
m = 12.0 g (given)
c1 = 2.09 J/(g°C) (given)
ΔT1 = 0°C - (-12.0°C) = 12.0°C
Q1 = 12.0 g * 2.09 J/(g°C) * 12.0°C
2. Heat required to melt the ice at 0°C:
Q2 = m * ΔH_fusion
ΔH_fusion is the heat of fusion (or melting) for ice, which is 333.5 J/g (a constant value)
Q2 = 12.0 g * 333.5 J/g
3. Heat required to warm the melted ice (now water) to its boiling point (100°C):
Q3 = m * c2 * ΔT2
c2 = 4.18 J/(g°C), which is the specific heat capacity of water
ΔT2 = 100°C - 0°C = 100°C
Q3 = 12.0 g * 4.18 J/(g°C) * 100°C
4. Heat required to vaporize the water at its boiling point (100°C):
Q4 = m * ΔH_vaporization
ΔH_vaporization is the heat of vaporization for water, which is 2260 J/g (a constant value)
Q4 = 12.0 g * 2260 J/g
5. Heat required to warm the steam to the final temperature (111°C):
Q5 = m * c3 * ΔT3
c3 = 2.01 J/(g°C), which is the specific heat capacity of steam
ΔT3 = 111°C - 100°C = 11°C
Q5 = 12.0 g * 2.01 J/(g°C) * 11°C
To calculate the total amount of heat required, add up the Q values:
Total heat (Q_total) = Q1 + Q2 + Q3 + Q4 + Q5
Now, substitute the values and calculate the heat:
Q1 = 12.0 g * 2.09 J/(g°C) * 12.0°C
Q2 = 12.0 g * 333.5 J/g
Q3 = 12.0 g * 4.18 J/(g°C) * 100°C
Q4 = 12.0 g * 2260 J/g
Q5 = 12.0 g * 2.01 J/(g°C) * 11°C
Q_total = Q1 + Q2 + Q3 + Q4 + Q5
Once you calculate this expression, you will arrive at the correct answer for the amount of heat (in kilojoules) required to warm the ice to steam at the given temperatures.