equal masses of ice at 0 oC, water aat 50 oC and steam at 100 oC are mixed and allowed to reach equilibrium. what will be the final temp of the mixture? and what percentage is the water and what percent will be steam??
To determine the final temperature of the mixture, we can use the concept of heat transfer. The heat gained or lost by a substance can be calculated using the formula:
Q = mcΔT
Where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we have equal masses of ice, water, and steam. Let's assume the mass of each substance is 1 gram for simplicity.
1. Heat gained by melting the ice to reach 0°C:
Q1 = m * c_ice * ΔT_1
= 1g * 2.09 J/g°C * (0 - (-0))°C
= 0 J
As ice remains at 0°C, no change in temperature occurs.
2. Heat gained by raising the temperature of water from 50°C to 100°C:
Q2 = m * c_water * ΔT_2
= 1g * 4.18 J/g°C * (100 - 50)°C
= 209 J
3. Heat lost by cooling down steam from 100°C to the final temperature:
Q3 = m * c_steam * ΔT_3
= 1g * 2.03 J/g°C * (100 - T_final)°C
= 203 J * (100 - T_final)
To reach thermal equilibrium, the heat gained from the water (Q2) and the heat lost from the steam (Q3) must be equal:
Q2 = Q3
209 J = 203 J * (100 - T_final)
Simplifying the equation, we get:
100 - T_final = 209/203 ≈ 1.03
T_final ≈ 100 - 1.03
T_final ≈ 98.97°C
Therefore, the final temperature of the mixture will be approximately 98.97°C.
Now let's calculate the percentage of water and steam in the mixture:
The total mass of the mixture is 1 gram + 1 gram + 1 gram = 3 grams.
The mass percentage of water can be calculated using the formula:
% water = (mass water / total mass) * 100
= (1g / 3g) * 100
≈ 33.33%
The mass percentage of steam can be calculated similarly:
% steam = (mass steam / total mass) * 100
= (1g / 3g) * 100
≈ 33.33%
Hence, the water and steam each make up approximately 33.33% of the mixture.