1.

what would be the final temperature if 10 g of steam at hundred degree celsius is mixed with 50 grams of ice at 0 degree Celsius ​

Answer»
  • To cool 50 g of water from 50 °C to 0 °C WOULD REQUIRE the removal of 4.2 x 50 x 50 =10500 J.

  • To MELT the ice would require the addition of 334 x 10 = 3340 J

  • 10500 > 3340 thus you can melt all the ice and have some heat to SPARE, specifically 10500 - 3340 = 7160 J

  • Now use this to warm up 10 + 50 = 60 g of water at 0 °C

  • 7160 / (60 x 4.2) = 28.4 °C


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