In an energy recycling process, X g of steam at 100 °C becomes water at 100 °C which converts Y g of ice at 0 °C into water at 100 °C. The ratio of X and Y will be

Let’s analyze the energy transfer to solve this problem.

Step 1: Energy released by X g of steam
When steam at 100°C condenses to water at 100°C, the energy released is:
Qₛₜₑₐₘ = X ⋅ Lᵥ
where Lᵥ = 540 cal/g (latent heat of vaporization of water).

Step 2: Energy absorbed by Y g of ice
The ice first melts into water at 0°C, and then this water is heated to 100°C. The total energy absorbed by Y g of ice is:
Qᵢcₑ = Y ⋅ Lf + Y ⋅ c ⋅ ΔT

where:
– Lf = 80 cal/g (latent heat of fusion of ice),
– c = 1 cal/g°C (specific heat capacity of water),
– ΔT = 100°C.

Put values:
Qᵢcₑ = Y ⋅ 80 + Y ⋅ 1 ⋅ 100 = Y ⋅ (80 + 100) = Y ⋅ 180 cal.

Step 3: Energy conservation

The energy released by steam is equal to the energy absorbed by ice:
X ⋅ 540 = Y ⋅ 180

Step 4: Solve for X/Y
X / Y = 180 / 540 = 1 / 3

Final Answer:
The ratio of X and Y is 1:3.

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