25 kg of ice at 0° C is combined with 4 kg of steam at 100° C. What is the final equilibrium temperature in ° C of the system? (Lf = 3.33 x 105 J/kg; Lv 2.26 x 106 J/kg, specific heat capacity of water =4186 J/(kg.K))

Question 2 options:
A) 40

B) 20

C) 60

D) 100

E) 8

Well, it seems like these ice and steam are about to have a "hot and cool" party! Let's see what happens when they get together.

So, to find the final temperature, we need to consider the heat transferred between the ice and steam until they reach equilibrium.

For the ice, we can use the equation Q = m * Lf, where Q is the heat transferred, m is the mass of ice, and Lf is the latent heat of fusion. Plugging in the values, we get Q = 25 kg * 3.33 x 10^5 J/kg = 8.325 x 10^6 J.

Now, let's consider the steam. We can use the equation Q = m * Lv, where Q is the heat transferred, m is the mass of steam, and Lv is the latent heat of vaporization. Plugging in the values, we get Q = 4 kg * 2.26 x 10^6 J/kg = 9.04 x 10^6 J.

Since heat lost by the steam is equal to heat gained by the ice (assuming no other heat transfer), we can equate the two equations:

8.325 x 10^6 J = 9.04 x 10^6 J

But wait, that's not true! Looks like we have a contradiction here. I guess these ice and steam aren't the best of friends after all. They just can't agree on a final temperature.

So, in this case, there is no equilibrium temperature that can be reached. Looks like the party got canceled. Sorry to be the bearer of bad news.

To find the final equilibrium temperature of the system, we need to calculate the amount of heat exchanged between the ice and the steam.

First, let's calculate the heat gained by the ice:

Q_ice = m_ice * L_f

where Q_ice is the heat gained by the ice, m_ice is the mass of the ice, and L_f is the latent heat of fusion.

Q_ice = 25 kg * 3.33 x 10^5 J/kg
= 8.33 x 10^6 J

Next, let's calculate the heat lost by the steam:

Q_steam = m_steam * L_v

where Q_steam is the heat lost by the steam, m_steam is the mass of the steam, and L_v is the latent heat of vaporization.

Q_steam = 4 kg * 2.26 x 10^6 J/kg
= 9.04 x 10^6 J

Now, let's calculate the heat gained by the ice as it reaches its final equilibrium temperature:

Q_ice_water = m_ice * c_water * (T_f - 0°C)

where Q_ice_water is the heat gained by the ice as it warms up to the final equilibrium temperature, m_ice is the mass of the ice, c_water is the specific heat capacity of water, and T_f is the final equilibrium temperature.

Q_ice_water = 25 kg * 4186 J/(kg.K) * (T_f - 0°C)
= 104650 J/(kg.K) * T_f

Next, let's calculate the heat lost by the steam as it cools down to the final equilibrium temperature:

Q_steam_water = m_steam * c_water * (100°C - T_f)

where Q_steam_water is the heat lost by the steam as it cools down to the final equilibrium temperature, m_steam is the mass of the steam, c_water is the specific heat capacity of water, and T_f is the final equilibrium temperature.

Q_steam_water = 4 kg * 4186 J/(kg.K) * (100°C - T_f)
= 16744 J/(kg.K) * (100°C - T_f)

According to the principle of conservation of energy, the sum of the heat gained and lost by the ice and steam should be zero:

Q_ice + Q_ice_water + Q_steam + Q_steam_water = 0

8.33 x 10^6 J + 104650 J/(kg.K) * T_f + 9.04 x 10^6 J + 16744 J/(kg.K) * (100°C - T_f) = 0

Simplifying the equation:

8.33 x 10^6 J + 104650 J/(kg.K) * T_f + 9.04 x 10^6 J + 16744 J/(kg.K) * 100°C - 16744 J/(kg.K) * T_f = 0

17.37 x 10^6 J + 8806 J/(kg.K) * T_f - 16744 J/(kg.K) * T_f = 0

(8806 J/(kg.K) - 16744 J/(kg.K)) * T_f = -17.37 x 10^6 J - 8806 J/(kg.K) * 100°C

(8806 J/(kg.K) - 16744 J/(kg.K)) * T_f = -17.37 x 10^6 J - 8806 J/(kg.K) * 100°C

(14866 J/(kg.K)) * T_f = -17.37 x 10^6 J - 880600 J/(kg.K)

T_f = (-17.37 x 10^6 J - 880600 J/(kg.K)) / (14866 J/(kg.K))

T_f = (-17.37 x 10^6 J) / (14866 J/(kg.K)) - (880600 J/(kg.K)) / (14866 J/(kg.K))

T_f = -1168.05 K + (-59.15 K)

T_f = -1227.2 K

Since the temperature cannot be negative, the final equilibrium temperature is 0°C.

Therefore, the answer is not provided in the options.

To find the final equilibrium temperature of the system, we need to calculate the heat gained or lost by each substance during the process.

First, let's calculate the heat gained or lost by the ice to reach the final equilibrium temperature. We can use the formula Q = m * Lf, where Q is the heat gained or lost, m is the mass, and Lf is the latent heat of fusion.

The mass of the ice is 25 kg and the latent heat of fusion is 3.33 x 10^5 J/kg. So the heat gained or lost by the ice is Q_ice = 25 kg * 3.33 x 10^5 J/kg.

Next, let's calculate the heat gained or lost by the steam to reach the final equilibrium temperature. We can use the formula Q = m * Lv, where Q is the heat gained or lost, m is the mass, and Lv is the latent heat of vaporization.

The mass of the steam is 4 kg and the latent heat of vaporization is 2.26 x 10^6 J/kg. So the heat gained or lost by the steam is Q_steam = 4 kg * 2.26 x 10^6 J/kg.

Now, let's calculate the heat gained or lost by the water formed by the ice and steam. We can use the formula Q = m * c * ΔT, where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The specific heat capacity of water is 4186 J/(kg.K). Since the mass of the water formed is the sum of the masses of the ice and steam (25 kg + 4 kg), and the initial and final temperatures are 0°C and the final equilibrium temperature respectively, the heat gained or lost by the water is Q_water = (25 kg + 4 kg) * 4186 J/(kg.K) * ΔT.

Since the total heat gained or lost by the system must be zero at equilibrium, we can set up the equation: Q_ice + Q_steam + Q_water = 0.

Solving for the final equilibrium temperature, we get:

25 kg * 3.33 x 10^5 J/kg + 4 kg * 2.26 x 10^6 J/kg + (25 kg + 4 kg) * 4186 J/(kg.K) * ΔT = 0.

Now, let's solve the equation to find the final equilibrium temperature, ΔT.

25 kg * 3.33 x 10^5 J/kg + 4 kg * 2.26 x 10^6 J/kg + (25 kg + 4 kg) * 4186 J/(kg.K) * ΔT = 0.

Simplifying the equation, we get:

83.25 x 10^5 J + 9.04 x 10^6 J + 29 kg * 4186 J/(kg.K) * ΔT = 0.

Combining the terms and solving for ΔT, we get:

ΔT = -(83.25 x 10^5 J + 9.04 x 10^6 J) / (29 kg * 4186 J/(kg.K)).

Calculating the value, we find ΔT ≈ -25.82°C.

Since we are looking for the final equilibrium temperature, we add ΔT to the initial temperature of 0°C:

Final equilibrium temperature = 0°C + (-25.82°C) ≈ -25.82°C.

Since a negative temperature is not physically meaningful in this context, we can conclude that the final equilibrium temperature of the system is approximately 0°C.

Therefore, none of the options (A) 40, (B) 20, (C) 60, (D) 100, or (E) 8 is the correct answer.

If all the ice melts, 25*Lf = 8.33*10^6 kJ is absorbed in that process. If all of the steam condenses, 4*Lv = 9.04*10^6 kJ is released in that process. There is enough steam to melt all the ice.

0.71*10^6 J are still available from condensation of remaining to raise the temperature of the melted ice. To raise all of the 25 kg to 100 C would require 1.047*10^7 kJ. So the melted water will be raised to an intermediate temperature between 0 and 100 C.

Calculate the final temperature by assuming all of the steam condenses and is lowered to final temperature Tf, releasing heat that melts all the ice and raises it to the same Tf.

25*(3.33*10^5 + 4186 Tf) = 4*(2.26*10^6 +4*(100 - Tf)]

Solve for Tf

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