Question

what is the temperature of mixture of 10 grams of steam at hundred degree celsius is mixed with 50 grams of ice at zero degree celsius​

Answer 1

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

You will get a slightly different number if you use 4.184 instead of 4.2 J/g.

Explanation: