A 0.200 kg block of ice at �15ºC is placed
into a pan on a stove, heated to a liquid, and
then to vapour with a final temperature of
115ºC. Calculate the total amount of heat
required for this process.
Well, what a transformation! From a solid block of ice to a liquid and then to a vapor. This ice really knows how to go with the flow! But let's get down to the calculations.
To determine the total amount of heat required for this process, we need to consider the three stages individually:
1. Heating the ice from -15ºC to 0ºC:
We can use 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. Since the ice is going from a solid to a liquid, we use the specific heat capacity of ice, which is 2100 J/kg°C.
Q = (0.200 kg) * (2100 J/kg°C) * (0°C - (-15°C))
2. Melting the ice at 0ºC:
When the ice reaches 0ºC, we need to consider the heat required for the phase change, known as the latent heat of fusion. For ice, it is 334,000 J/kg.
Q = (0.200 kg) * (334,000 J/kg)
3. Heating the liquid water from 0ºC to 115ºC:
Once the ice has turned into water, we can use the same formula as before since it's a pure liquid. Let's use the specific heat capacity of water, which is 4200 J/kg°C, shall we?
Q = (0.200 kg) * (4200 J/kg°C) * (115°C - 0°C)
Now, my calculating circuits need a moment to compute all these numbers. Beep, boop, beep... Ah, there we go! Adding up the results of each of the three stages will give us the total amount of heat required for this chilly, then toasty, then steamy journey.
To calculate the total amount of heat required for this process, we need to consider the three different phases of the substance: solid, liquid, and vapor. We will calculate the heat required for each phase separately and then sum them up to get the total heat required.
First, we need to calculate the heat required to raise the temperature of the ice from -15ºC to its melting point, which is 0ºC. This is given by the equation:
Q1 = mcΔT
Where:
m = mass of the ice = 0.200 kg
c = specific heat capacity of ice = 2,093 J/kgºC
ΔT = change in temperature = 0ºC - (-15ºC) = 15ºC
Substituting the values into the equation, we get:
Q1 = (0.200 kg)(2,093 J/kgºC)(15ºC) = 6,279 J
Next, we need to calculate the heat required to melt the ice at 0ºC into liquid water at 0ºC. This is given by the equation:
Q2 = mLf
Where:
m = mass of the ice = 0.200 kg
Lf = latent heat of fusion of ice = 334,000 J/kg (or 334 kJ/kg)
Substituting the values into the equation, we get:
Q2 = (0.200 kg)(334,000 J/kg) = 66,800 J
Finally, we need to calculate the heat required to raise the temperature of the liquid water from 0ºC to 115ºC. This is given by the equation:
Q3 = mcΔT
Where:
m = mass of the water = 0.200 kg
c = specific heat capacity of water = 4,186 J/kgºC (or 4.186 kJ/kgºC)
ΔT = change in temperature = 115ºC - 0ºC = 115ºC
Substituting the values into the equation, we get:
Q3 = (0.200 kg)(4,186 J/kgºC)(115ºC) = 96,278 J
Now, we can sum up the heat required for each phase:
Total heat required = Q1 + Q2 + Q3
= 6,279 J + 66,800 J + 96,278 J
= 169,357 J
Therefore, the total amount of heat required for this process is approximately 169,357 J.
To calculate the total amount of heat required for this process, we need to consider three stages:
1. Heating the ice from -15ºC to 0ºC (melting)
2. Heating the water from 0ºC to 100ºC (boiling)
3. Heating the steam from 100ºC to 115ºC
Let's break down the calculation for each stage:
1. Melting the Ice:
The heat required to melt the ice can be calculated using the formula:
Q = m * Lf
Where:
Q is the heat required (in joules)
m is the mass of the ice (in kilograms)
Lf is the latent heat of fusion for water (in joules per kilogram)
The latent heat of fusion for water is 334,000 J/kg.
Q1 = 0.200 kg * 334,000 J/kg = 66,800 J
Therefore, it requires 66,800 J of heat to melt the ice.
2. Heating the Water:
The heat required to heat the water from 0ºC to 100ºC can be calculated using the formula:
Q = m * Cp * ΔT
Where:
Q is the heat required (in joules)
m is the mass of the water (in kilograms)
Cp is the specific heat capacity of water (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
The specific heat capacity of water is approximately 4,186 J/(kg·°C).
ΔT2 = 100°C - 0°C = 100°C
Q2 = 0.200 kg * 4,186 J/(kg·°C) * 100°C = 83,720 J
Therefore, it requires 83,720 J of heat to heat the water.
3. Heating the Steam:
The heat required to heat the steam from 100ºC to 115ºC can be calculated using the formula:
Q = m * Cp * ΔT
Where:
Q is the heat required (in joules)
m is the mass of the steam (in kilograms)
Cp is the specific heat capacity of steam (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
The specific heat capacity of steam is approximately 2,000 J/(kg·°C).
ΔT3 = 115°C - 100°C = 15°C
Q3 = 0.200 kg * 2,000 J/(kg·°C) * 15°C = 6,000 J
Therefore, it requires 6,000 J of heat to heat the steam.
Total Heat Required:
To calculate the total amount of heat required for the entire process, we simply add up the heat values from each stage:
Total Heat = Q1 + Q2 + Q3
Total Heat = 66,800 J + 83,720 J + 6,000 J
Total Heat = 156,520 J
Therefore, the total amount of heat required for this process is 156,520 J.
If it started out at 15 C, it wasn't ice. Did you leave out a minus sign?
You are going to have to look up the specific heats of ice, water and steam, the heat of vaporization, and the heat of fusion. If you still need help, try posting again.