0.02kg of ice and 0.1kg of water at 0celsius are in a container. Steam at 100celsuis is passed in until all the ice just melted. How much water is now in the container?

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To determine the amount of water remaining in the container after melting all the ice, we need to calculate the energy required to melt the ice and the resulting increase in temperature of the water.

First, let's calculate the energy required to melt the ice:

The specific heat capacity of ice is 2.09 J/g°C.
The latent heat of fusion (energy required to melt ice) is 334 J/g.

The mass of ice is 0.02 kg (20 g).
The initial temperature of the ice is 0°C.

Energy required to melt the ice = mass of ice × latent heat of fusion
= 20 g × 334 J/g
= 6680 J

Since all the energy provided by the steam will be used to melt the ice, this energy is also the energy the steam releases.

Now, let's calculate the increase in temperature of the resulting water:

The specific heat capacity of water is 4.18 J/g°C.

The mass of water is 0.1 kg (100 g).
The final temperature of the water is the same as the boiling point of steam, which is 100°C.

Energy required to increase the temperature of water = mass of water × specific heat capacity × change in temperature
= 100 g × 4.18 J/g°C × (100°C - 0°C)
= 41800 J

Now, let's calculate the total energy released by the steam:

Total energy released by the steam = Energy required to melt the ice + Energy required to increase the temperature of the water
= 6680 J + 41800 J
= 48480 J

Finally, let's determine the mass of water in the container after all the ice has melted:

Mass of water = Total energy released by the steam ÷ Latent heat of vaporization of water
= 48480 J ÷ 2260 J/g
≈ 21.46 g

Therefore, there will be approximately 21.46 grams of water remaining in the container after all the ice has melted.

To find out how much water is in the container after the ice has melted, we need to consider the heat transferred from the steam to melt the ice and then heat the water.

First, let's calculate the heat required to melt the ice. The specific heat capacity of ice is 2.09 J/g°C, and the heat equation is:

Q = m × c × ΔT

Where:
Q = heat transferred (in joules)
m = mass (in grams)
c = specific heat capacity (in joules/gram°C)
ΔT = change in temperature (in °C)

Given:
Mass of ice (m1) = 0.02 kg = 20 grams
Specific heat capacity of ice (c1) = 2.09 J/g°C
Temperature change to melt the ice (ΔT1) = 0°C (ice at 0°C)

Q1 = m1 × c1 × ΔT1
Q1 = 20 g × 2.09 J/g°C × 0°C
Q1 = 0 J

The heat required to melt the ice is 0 joules since there is no temperature change.

Next, let's calculate the heat required to heat the water. The specific heat capacity of water is 4.18 J/g°C.

Given:
Mass of water (m2) = 0.1 kg = 100 grams
Specific heat capacity of water (c2) = 4.18 J/g°C
Temperature change needed for water (ΔT2) = 100°C (from 0°C to 100°C)

Q2 = m2 × c2 × ΔT2
Q2 = 100 g × 4.18 J/g°C × 100°C
Q2 = 418,000 J

The heat required to heat the water is 418,000 joules.

Now, let's add the heat required to melt the ice and heat the water to find the total heat transferred.

Total heat transferred (Q_total) = Q1 + Q2
Q_total = 0 J + 418,000 J
Q_total = 418,000 J

Assuming no heat is lost to the surroundings, this heat is supplied by the steam. The heat of vaporization for water is 2,260 J/g. Let's find the mass of the steam required.

Q_total = m3 × L
418,000 J = m3 × 2,260 J/g
m3 = 418,000 J / 2,260 J/g
m3 ≈ 185 grams

So, approximately 185 grams of steam is required to melt the ice and heat the water.

Now, let's determine the final mass of water in the container by subtracting the mass of the initially present water from the final total mass.

Final mass of water = (Mass of water initially present) + (Mass of steam added) - (Mass of ice initially present)
Final mass of water = 100 g + 185 g - 0 g
Final mass of water = 285 g

Therefore, after all the ice has melted and the steam has been added, there will be 285 grams of water in the container.

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