Question

Thermal Physics 2. A secondary steam is passed into a well- lagged copper can of mass 250 g containing 400 g of water and 50 g of ice at 0°C. The mixture is well stirred and the steam supplied cut off when the temperature of the can and its contents reach 20 °C. Neglecting heat losses find the mass of the steam condensed. SHC of water - 4200 J kg K SCH of copper - 400 J kg 1, of steam - 2260000 J kg Lofice-336000 J kg K​

Answer 1

The mass of the condensed steam is approximately 0.0003765 kg or 0.3765 grams.

Step-by-step explanation:

To find the mass of the steam condensed, we need to calculate the total heat gained by the water, ice, and copper can, and then equate it to the heat lost by the steam during the condensation process.

Let's denote:

m1 = mass of water (400 g)

m2 = mass of ice (50 g)

m3 = mass of copper can (250 g)

m4 = mass of steam to be determined (let's assume it's in kg)

Specific heat capacities:

SHC_water = 4200 J/(kgK)

SHC_copper = 400 J/(kgK)

SHC_steam = 2260000 J/(kgK)

SHC_ice = 336000 J/(kgK)

The heat gained by the water to reach 20°C from 0°C:

Q1 = m1 * SHC_water * (20°C - 0°C)

The heat gained by the ice to reach 20°C from 0°C:

Q2 = m2 * SHC_ice * (20°C - 0°C)

The heat gained by the copper can to reach 20°C from 0°C:

Q3 = m3 * SHC_copper * (20°C - 0°C)

Now, let's calculate the heat lost by the steam during condensation. The heat lost by the steam is equal to the heat gained by the water, ice, and copper can:

Q_lost = Q1 + Q2 + Q3

When steam condenses, it releases latent heat, which is given by:

Latent heat of steam (L) = 2260000 J/kg

Therefore, the total heat lost by the steam (Q_lost) is also used to raise its temperature from 100°C (boiling point) to 20°C:

Q_lost = m4 * SHC_steam * (100°C - 20°C) + m4 * L

Now we can equate the two expressions for Q_lost:

m4 * SHC_steam * (100°C - 20°C) + m4 * L = Q1 + Q2 + Q3

Now let's plug in the values and solve for m4:

m4 * 2260000 * 80 + m4 * 2260000 = m1 * 4200 * 20 + m2 * 336000 * 20 + m3 * 400 * 20

Now, let's calculate the masses in kg:

m1 = 400 g = 0.4 kg

m2 = 50 g = 0.05 kg

m3 = 250 g = 0.25 kg

Now, solve for m4:

m4 * 181600000 + m4 * 2260000 = 0.4 * 4200 * 20 + 0.05 * 336000 * 20 + 0.25 * 400 * 20

m4 * (181600000 + 2260000) = 33600 + 33600 + 2000

m4 * 183860000 = 69200

m4 = 69200 / 183860000

m4 ≈ 0.0003765 kg

So, the mass of the condensed steam is approximately 0.0003765 kg or 0.3765 grams.