Question

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 1

• 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

Answer 2

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There are two things - 10 g of steam at 100 deg and 50 g of ice at 0 deg.

Latent heat of ice is 80 cal/g and for steam, it is 540 cal/g

The 10 g steam at 100 deg. release (10*540 cal =) 5400 cal heat and become 10 g of water at 100 deg.

The 50 g ice at 0 deg. absorb (50*80 cal =) 4000 cal heat and become 50 g of water at 0 deg.

The heat released by the steam is greater than the heat absorbed by the ice. And the released heat must be equal to the absorbed heat.

So, 1st ice absorbs 4000 cal and converts into the water. Then the rest 1400 cal heat increases the temperature of water from 0 deg to t Kelvin (assume) and with this 10 g of steam converts into the water of 100 deg.

The specific heat of water is 1 cal/g deg-Celcius.

Heat balance equation:

50 * 1 * (t-273) = 1400

t = 28 + 273 = 301 K

Now, the mixture contains 50 g of water at 301 K (= 28 deg) and 10 g of water at 100 deg (373 K)

Assume, the final temperature of the mixture is T K

Heat balance equation:

50 * 1 * (T- 301) = 10 * 1 * (373 -T)

50 T - 15050 = 3730 -10 T

60 T = 3730 + 15050 = 18780

T = 313 K

So, the final mixture will be 60 g of water at (313 K =) 40 deg.