1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10³ J kg⁻¹ and latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹.

Open in App

Solution

Given:
Amount of ice at 0^oC = 1 kg
Amount of steam at 100^oC = 1 kg
Latent heat of fusion of ice = 3.36 × 10³ J kg⁻¹
Latent heat of vapourisation of water = 2.26 × 10⁶ J kg⁻¹

We can observe that the latent heat of fusion of ice (3.36 × 10⁵J kg⁻¹) is smaller that latent heat of vapouisation of water (2.26 × 10⁶ ). Therefore, ice will first change into water as less heat is required for this and there will be equilibrium between steam and water.

Heat absorbed by the ice when it changes into water (Q₁) = 1×(3.36 × 10⁵) J
Heat absorbed by the water formed to change its temperature from 0^oC to 100^oC (Q₂) = 1 × 4200 × 100 = 4.2 × 10⁵ J
Total heat absorbed by the ice to raise the temperature to 100°C, Q = Q₁+Q₂ = 3.36 × 10⁵ + 4.2 × 10⁵ = (3.36 + 4.2) × 10⁵ = 7.56 × 10⁵ J

The heat required to change ice into water at 100^oC is supplied by the steam. This heat will be released by the steam and will then change into water.
If all the steam gets converted into water, heat released by steam, Q' = 1 ×( 2.26 × 10⁶) J = 2.26 × 10⁶ J
Amount of heat released is more than that required by the ice to get converted into water at 100^oC. Thus,
Extra heat = Q − Q'
= (2.26 − 0.756) × 10⁶
= 1.506 × 10⁶

Let the mass of steam that is condensed into water be m. Thus,
m $= \frac{7.56 \times 10^{5}}{2.26 \times 10^{6}}$ = 0.335 kg = 335 gm

Total amount of water at 100°C = 1000 + 335 = 1335 g =1.335 g
Steam left = 1− 0.335 = 0.665 kg = 665 gm

Suggest Corrections

Similar questions

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36 \times 10^{5} J k g^{- 1}$ and latent heat of vaporization of water = $2.26 \times 10^{6} J k g^{- 1}$

1 k g

of ice at

0^{\circ} C

is mixed with

1 k g

of steam at

100^{\circ} C

. What will be the composition of the system when thermal equilibrium is reached? Latent of fusion

3.36 \times 10 J / g

and latent heat of vaporization of water =

2.26 \times 10 J / g

1 kg

of ice at

0^{\circ} C

is mixed with

1 kg

of steam at

100^{\circ} C

. Find the mass of steam when the mixture has reached thermal equilibrium.

[L_{f u s i o n, i c e} = 3.36 \times 10^{5} J/kg,

L_{v a p o r i z a t i o n, w a t e r} = 2.26 \times 10^{6} J/kg,

C_{w a t e r} = 4200 J/kg-K]

Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ , latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = 2.25 × 10⁶ J kg⁻¹

Join BYJU'S Learning Program

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 103 J kg−1 and latent heat of vaporization of water = 2.26 × 106 J kg−1.

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10³ J kg⁻¹ and latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹.