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

To calculate 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 three main steps involved:

1. Heating the ice from -30°C to 0°C
2. Melting the ice at 0°C
3. Heating the water from 0°C to 100°C and then vaporizing it at 100°C

Let's break down the calculation for each step:

1. Heating the ice from -30°C to 0°C:
The heat required to raise the temperature of 0.1 g of ice by 1°C can be calculated using the specific heat capacity formula:
Q = m * c * ΔT

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

For the ice, we have:
m = 0.1 g
c = 2 J/g · °C
ΔT = (0°C - (-30°C)) = 30°C

So, the heat required to heat the ice from -30°C to 0°C is:
Q1 = 0.1 g * 2 J/g · °C * 30°C = 6 J

2. Melting the ice at 0°C:
The heat required to melt the ice can be calculated using the latent heat of fusion formula:
Q = m * Hfus

Where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
Hfus = latent heat of fusion (in J/g)

For the ice, we have:
m = 0.1 g
Hfus = 340 J/g

So, the heat required to melt the ice at 0°C is:
Q2 = 0.1 g * 340 J/g = 34 J

3. Heating the water from 0°C to 100°C and then vaporizing it at 100°C:
The heat required to raise the temperature of 0.1 g of water by 1°C can be calculated using the specific heat capacity formula:
Q = m * c * ΔT

For the water, we have:
m = 0.1 g
c = 4 J/g · °C
ΔT = (100°C - 0°C) = 100°C

So, the heat required to heat the water from 0°C to 100°C is:
Q3 = 0.1 g * 4 J/g · °C * 100°C = 40 J

After reaching 100°C, we need to consider the heat required for vaporization. The heat required to vaporize 0.1 g of water can be calculated using the latent heat of vaporization formula:
Q = m * Hvap

For the water, we have:
m = 0.1 g
Hvap = 2260 J/g

So, the heat required to vaporize the water at 100°C is:
Q4 = 0.1 g * 2260 J/g = 226 J

Finally, we can calculate the total heat required by summing up the individual heats for each step:
Total heat = Q1 + Q2 + Q3 + Q4 = 6 J + 34 J + 40 J + 226 J = 306 J

Therefore, approximately 306 joules of heat is required to heat 0.1 g of ice at -30°C to steam at 100°C.

See your earlier post.