NEETPhysicsProperties of MatterMCQ+4 / −1
Steam at 100oC is passed into 20 g of water at 10oC. When water acquires a temperature of 80oC, the mass of water present will be
[Take specific heat of water = 1 cal g1 oC1 and latent heat of steam = 540 cal g1]
[Take specific heat of water = 1 cal g1 oC1 and latent heat of steam = 540 cal g1]
- A24 g
- B31.5 g
- C42.5 g
- D22.5 g
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Correct answer: D
According to the principle of calorimetry.
Heat lost = Heat gained
m × 540 + m × 1 × (100 – 80)
= 20 × 1 × (80 – 10)
m = 2.5 g
Therefore total mass of water at 80°C
= (20 + 2.5) g = 22.5 g
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