If 1g of steam is mixed wid 1g of ice, then what will be the resultant temperature of the mixture?

To determine the resultant temperature of the mixture, we need to consider the heat transfer that occurs when steam condenses into water and when ice melts into water.

Step 1: Determine the amount of heat released when 1g of steam condenses into water.
The heat released during the condensation of steam can be calculated using the formula Q = m × L, where Q is the heat energy, m is the mass, and L is the latent heat of condensation. The latent heat of condensation for water is approximately 540 calories per gram.
Therefore, the heat released by 1g of steam is Q1 = 1g × 540 cal/g.

Step 2: Determine the amount of heat absorbed when 1g of ice melts into water.
Similar to step 1, the heat absorbed during the melting of ice can be calculated using the formula Q = m × L, where Q is the heat energy, m is the mass, and L is the latent heat of fusion. The latent heat of fusion for water is approximately 80 calories per gram.
Therefore, the heat absorbed by 1g of ice is Q2 = 1g × 80 cal/g.

Step 3: Calculate the net heat transfer.
The net heat transfer is given by Q = Q1 - Q2, where Q1 is the heat released by the steam and Q2 is the heat absorbed by the ice.
Q = (1g × 540 cal/g) - (1g × 80 cal/g) = 460 cal.

Step 4: Calculate the resulting temperature.
The heat transferred is also equal to the change in temperature multiplied by the specific heat capacity of water, which is approximately 1 calorie per gram per degree Celsius.
So, Q = m × ΔT × c, where Q is the heat transferred, m is the mass, ΔT is the change in temperature, and c is the specific heat capacity of water.
Rearranging the formula, we get ΔT = Q / (m × c).

Substituting the values, we have ΔT = 460 cal / (2g × 1 cal/g°C) = 230°C.

Therefore, the resulting temperature of the mixture will be 230°C.

To find the resultant temperature of the mixture, we need to consider the heat exchange that occurs when steam and ice are mixed.

1. First, let's calculate the amount of heat released or absorbed by each substance.

- Heat released by steam: The heat released when steam condenses into water can be calculated using the formula: Q = m × L, where Q is the heat energy, m is the mass, and L is the specific latent heat of vaporization. For water, the specific latent heat of vaporization is 2260 Joules per gram.
Q_steam = 1g × 2260 J/g = 2260 J

- Heat absorbed by ice: The heat absorbed when ice melts into water can be calculated using the same formula: Q = m × L, where L is the specific latent heat of fusion. For water, the specific latent heat of fusion is 334 J/g.
Q_ice = 1g × 334 J/g = 334 J

2. Next, let's determine the direction of heat flow.

Since heat always flows from a higher temperature to a lower temperature until thermal equilibrium is reached, the heat released by the steam will be absorbed by the ice until both substances reach the same temperature.

3. Now, let's find the temperature at which the heat exchange occurs.

To find the resultant temperature, we can use the principle of energy conservation, which states that the total heat released is equal to the total heat absorbed:

Q_steam = Q_ice
2260 J = 334 J + mL × ΔT

Since the mass of steam and ice is the same (1g), we can cancel it out:

2260 J = 334 J + L × ΔT

Rearranging the equation to solve for ΔT:

ΔT = (2260 J - 334 J) / L

Plugging in the value for L (specific latent heat of fusion or vaporization), we find:

ΔT = (2260 J - 334 J) / 334 J/g = 5.68 degrees Celsius

Therefore, the resultant temperature of the mixture will be 5.68 degrees Celsius.

can there be ne other answers elena??????????

We start with the assumption that all the steam condenses and all the ice melts

.r•m+ c•m• (t-0) = λ•m + c•m• (100 – t),
where r = 334000 J/kg is the heat of fusion,
c =4180 J/kg•K,
λ = 2250000J/kg is the heat of vaporization.
Since the mass of steam = the mass of ice,
r+ c•(T-0) = λ + c•(100 – T),
Solve for T
T =279.2 oC
Since this is impossible, the initial assumption was false, the steam does not completely condense, and the ice heats up to 100 *C (this is the answer, as all matter in the system is at 100 *C)

To see how much steam condenses, we have 2250000•x=334000+4180 • 100,
where x is the proportion of steam that condenses.

x =0.334

0.334g of steam condense, since this is physically possible this is a successful check on the answer.