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Poll

Steam at 100°C is passed into 20 g of water at 10°C, then water acquires a temperating of 80°C, the man of water present will be [Take specific heat of water = 1 cal g⁻¹°C⁻¹and Latent heat of steam = 540 cal g⁻¹]

  • 12

Poll Results

0%(a) 24 g
0% (b) 31.5 g
0%(c) 42.5 g
100%(d) 22.5 g ( 14 voters )
Based On 14 Votes

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