How much energy is required to bring 10.0 grams of ice at -5.00 degrees to 105-degree steam? Assume the system is isolated so that no mass is lost to the surroundings

nobody knows because you did not give me:

specific heat of ice
heat of fusion of water
specific heat of water
heat of vaporization of water
Cp of steam

Energy =
10 grams (or .010 kg if using kg)
times the sum of:
[5 deg * spec heat ice
+ heat of fusion of water
+ 100 deg * spec heat of water
+ heat of vaporaization of water
+ 5 deg * specific heat of steam at one atmosphere ]

To calculate the amount of energy required to bring ice to steam, we need to consider three steps:

1. Heating the ice from its initial temperature of -5.00 degrees Celsius to its melting point at 0 degrees Celsius.
2. Melting the ice into water at 0 degrees Celsius.
3. Heating the water from 0 degrees Celsius to its final temperature of 100 degrees Celsius (assuming atmospheric pressure).

Let's break down each step and calculate the energy required:

Step 1: Heating the ice from -5.00°C to 0°C (solid phase)
The energy required to heat a substance can be calculated using the formula:

Q = m * c * ΔT

Where:
Q = energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)

For ice, the specific heat capacity is approximately 2.09 J/g°C.

Q1 = 10.0 g * 2.09 J/g°C * (0°C - (-5.00°C))

Q1 = 10.0 g * 2.09 J/g°C * 5.00°C

Q1 = 104.5 J

Step 2: Melting the ice (solid to liquid phase)
The energy (also known as heat of fusion) required to melt a substance can be calculated using the formula:

Q = m * ΔHf

Where:
m = mass of the substance (in grams)
ΔHf = heat of fusion (in J/g)

For ice, the heat of fusion is approximately 333.5 J/g.

Q2 = 10.0 g * 333.5 J/g

Q2 = 3335 J

Step 3: Heating the water from 0°C to 100°C (liquid phase)
The energy required to heat the water can be calculated using the same formula as step 1:

Q = m * c * ΔT

For water, the specific heat capacity is approximately 4.18 J/g°C.

Q3 = 10.0 g * 4.18 J/g°C * (100°C - 0°C)

Q3 = 4180 J

Total energy required:
Q_total = Q1 + Q2 + Q3
Q_total = 104.5 J + 3335 J + 4180 J
Q_total = 7620.5 J

Therefore, the total amount of energy required to bring 10.0 grams of ice at -5.00 degrees Celsius to 105-degree steam is approximately 7620.5 Joules.