Calculate the total heat energy needed to convert 110g of ice at -10°C to steam at 110°C..?

Well, let me put on my "heaty" thinking cap for a moment.

To calculate the total heat energy needed to convert ice at -10°C to steam at 110°C, we need to consider a few stages. First, we need to warm the ice from -10°C to 0°C, then we need to melt the ice into water at 0°C, after that we need to heat the water from 0°C to 100°C, and finally we need to change the water into steam at 100°C and bring it up to 110°C.

Each stage requires a certain amount of energy. To determine this, we need to use the formula:

Q = m * c * ΔT

Where:
Q is the amount of heat energy in Joules,
m is the mass in grams,
c is the specific heat capacity (for ice, water, and steam),
ΔT is the change in temperature.

So, let's break it down:

1. Warming the ice from -10°C to 0°C:
Q1 = 110g * 2.09 J/g°C * (0 - (-10))°C

2. Melting the ice into water at 0°C:
Q2 = 110g * 334 J/g

3. Heating the water from 0°C to 100°C:
Q3 = 110g * 4.18 J/g°C * (100 - 0)°C

4. Changing the water into steam at 100°C:
Q4 = 110g * 2260 J/g

5. Heating the steam from 100°C to 110°C:
Q5 = 110g * 2.03 J/g°C * (110 - 100)°C

Finally, to get the total heat energy, we add up all the individual values:

Total heat energy = Q1 + Q2 + Q3 + Q4 + Q5

But hey, I'm not just a math bot, I'm a clown bot! So, let's leave it to a serious calculator to crunch those numbers for you.

To calculate the total heat energy needed to convert ice at -10°C to steam at 110°C, we need to consider three steps:

1. Heating the ice from -10°C to 0°C.
2. Melting the ice at 0°C.
3. Heating the water from 0°C to 100°C and then converting it into steam at 100°C.

Step 1: Heating the ice from -10°C to 0°C.
The specific heat capacity (C) of ice is 2.09 J/g°C. Therefore, the heat energy required to heat the ice can be calculated using the formula:

Q1 = mass × C × change in temperature

Q1 = 110g × 2.09 J/g°C × (0°C - (-10°C))

Q1 = 110g × 2.09 J/g°C × 10°C

Q1 = 2299 J

Step 2: Melting the ice at 0°C.
The heat of fusion (ΔHfus) for ice is 334 J/g. Therefore, the heat energy required to melt the ice can be calculated using the formula:

Q2 = mass × ΔHfus

Q2 = 110g × 334 J/g

Q2 = 36740 J

Step 3: Heating the water from 0°C to 100°C and converting it into steam at 100°C.
The specific heat capacity (C) of water is 4.18 J/g°C. Therefore, the heat energy required to heat the water can be calculated using the formula:

Q3 = mass × C × change in temperature

Q3 = 110g × 4.18 J/g°C × (100°C - 0°C)

Q3 = 110g × 4.18 J/g°C × 100°C

Q3 = 45980 J

Total Heat Energy (Q) = Q1 + Q2 + Q3 = 2299 J + 36740 J + 45980 J = 85019 J

Therefore, the total heat energy needed to convert 110g of ice at -10°C to steam at 110°C is 85019 J.

To calculate the total heat energy needed, we need to consider the different phases of the substance and the energy required for each phase change.

1. Heating the ice from -10°C to 0°C:
The heat energy needed can be calculated using the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
For ice, the specific heat capacity is approximately 2.09 J/g°C.
So, the heat energy required to heat the ice from -10°C to 0°C can be calculated as:
Q1 = (mass of ice) * (specific heat capacity of ice) * (change in temperature)
= 110 g * 2.09 J/g°C * (0°C - (-10°C))
= 110 g * 2.09 J/g°C * 10°C
= 230.9 J

2. Melting the ice at 0°C:
The heat energy needed for the phase change from solid to liquid (melting) is calculated using the formula Q = m * ΔHf, where ΔHf is the heat of fusion.
For ice, the heat of fusion is approximately 334 J/g.
So, the heat energy required to melt the ice can be calculated as:
Q2 = (mass of ice) * (heat of fusion of ice)
= 110 g * 334 J/g
= 36,740 J

3. Heating the water from 0°C to 100°C:
The heat energy needed can be calculated using the same formula as before:
Q3 = (mass of water) * (specific heat capacity of water) * (change in temperature)
The specific heat capacity of water is approximately 4.18 J/g°C.
The change in temperature is 100°C - 0°C = 100°C.
So, the heat energy required to heat the water can be calculated as:
Q3 = (mass of water) * 4.18 J/g°C * 100°C

4. Vaporizing the water at 100°C:
The heat energy needed for the phase change from liquid to gas (vaporization) is calculated using the formula Q = m * ΔHv, where ΔHv is the heat of vaporization.
For water, the heat of vaporization is approximately 2260 J/g.
So, the heat energy required to vaporize the water can be calculated as:
Q4 = (mass of water) * 2260 J/g.

Now, to find the total heat energy required, we sum up all the individual values: Total heat energy = Q1 + Q2 + Q3 + Q4.

Substitute the values into the formulas mentioned above and calculate each part of the total heat energy required. Finally, sum all the heat energy values to get the total heat energy needed.

C=heat/mass*temperature

c=110/110*-10
c=110/-1100
c=0.1cal/g