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Step-by-Step Solution
Step 1: Identify All the Elements Involved
A calorimeter (with a water equivalent of 20 g) contains 180 g of water. Initial temperature of the water (and calorimeter) is 25°C. An unknown mass m of steam at 100°C is introduced, and the final mixture attains 31°C. We need to find m.
Step 2: Heat Gained by the Water in the Calorimeter
The 180 g of water (specific heat = 1 cal g–1°C–1) heats from 25°C to 31°C:
$ Q_{water} = 180 \times 1 \times (31 - 25) $
$ Q_{water} = 180 \times 6 \,=\, 1080 \text{ cal} $
Step 3: Heat Gained by the Calorimeter
The calorimeter has a water equivalent of 20 g, meaning it behaves like 20 g of water in terms of heating:
$ Q_{calorimeter} = 20 \times 1 \times (31 - 25) $
$ Q_{calorimeter} = 20 \times 6 \,=\, 120 \text{ cal} $
Step 4: Total Heat Gained by Water and the Calorimeter
Summing the heats from Steps 2 and 3:
$ Q_{gained} = 1080 + 120 = 1200 \text{ cal} $
Step 5: Heat Lost (Released) by Steam
Steam at 100°C first condenses to water at 100°C (releasing latent heat), then cools to 31°C. Let the steam mass be m.
1) Latent heat of condensation (steam to water), $ L = 540 \text{ cal g}^{-1} $.
2) Cooling from 100°C to 31°C involves a temperature change of (100 – 31) = 69°C.
(Specific heat of water = 1 cal g–1°C–1)
So total heat released by mass m of steam:
$ Q_{lost} = m \times 540 + m \times 1 \times (100 - 31) $
$ Q_{lost} = m \times 540 + m \times 69 = m \times (540 + 69) = m \times 609 $
Step 6: Equate Heat Gained to Heat Lost
Since no heat is lost to the surroundings ideally,
$ Q_{gained} = Q_{lost} \quad \Rightarrow \quad 1200 = 609\,m $
$ \Rightarrow m = \frac{1200}{609} \approx 1.97 \text{ g} $
Step 7: Conclude the Closest Answer
Rounding to the nearest option, m ≈ 2 g.
