Steam, at 1000C is passed into a container of negligible heat capacity, containing 20g of ice and 100g
of water at 00C, until the ice is completely melted. Determine the total mass of water in the container.
[Specific latent heat of steam = 2.3× l03
Jk-1
; specific latent heat of ice =3.4 × l02
Jg-1
; specific heat
capacity of water = 4.2Jg-1K
-1
]
To solve this problem, we need to consider the heat transfer that occurs during the melting of the ice and the heating of the water.
First, let's calculate the heat required to melt the ice:
Q1 = mass of ice * specific latent heat of ice
Q1 = 20g * 3.4 × 10^2 Jg^-1
Q1 = 6.8 × 10^3 J
Next, let's calculate the heat required to heat the water from 0°C to 100°C:
Q2 = mass of water * specific heat capacity of water * change in temperature
Q2 = 100g * 4.2 Jg^-1K^-1 * (100°C - 0°C)
Q2 = 4.2 × 10^5 J
Now, let's calculate the heat required to convert the water at 100°C into steam:
Q3 = mass of water * specific latent heat of steam
Q3 = mass of water * 2.3 × 10^3 Jg^-1
Q3 = 2.3 × 10^5 * mass of water
Since there is no heat loss or gain from the container, the total heat transferred is equal to the sum of Q1, Q2, and Q3:
Q_total = Q1 + Q2 + Q3
From the conservation of energy principle, the total heat transferred is equal to the heat absorbed by the steam:
Q_total = mass of steam * specific latent heat of steam
Since the mass of steam is equal to the mass of water in the container (assumed no mass loss due to evaporation),
mass of water = Q_total / specific latent heat of steam
mass of water = (Q1 + Q2 + Q3) / (2.3 × 10^3 Jg^-1)
Plugging in the values:
mass of water = (6.8 × 10^3 J + 4.2 × 10^5 J + 2.3 × 10^5 J) / (2.3 × 10^3 Jg^-1)
mass of water = 2.6 × 10^2 g
mass of water = 260 g
Therefore, the total mass of water in the container is 260 grams.