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Question

The question is 50g of steam at 165 degree is added to 200g of ice at -45 degree, what will be the final temperature of the system?

Answer 1

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you have to use the formula q=mc(final temp - initial temp)

The steam will lose heat (q) and therefore will be negative.

-q = mc(final temp - initial temp)

The ice will gain heat and will still be q = mc (final - initial).

The heat loss for the steam is equal to the heat gained from the ice so you can set the two formulas equal together, so...

- mc (final - initial) = mc (final - initial)

^^^^^ steam ^^^^^^^ ice

So now we can put in the numbers.

The specific heat (c) of steam is 1.87 J/g Celsius

The specific heat of ice is 2.06 J/g Celsius

The final temperature of the two will be the same.

-50 g * 1.87 J/gC (final-165C) = 200 g * 2.06 J/gC (final - (-45C))

Plug some numbers into your calculator and you get

-93.5 J/C (final-165C) = 412 J/C (final + 45C)

Divide both sides by 95.5 J/C and you get

final - 165C = -4.406 (final + 45 C)

Then distribute the -4.406 and you get

final -165 C = -4.046*final - 198.29 C

Combine the like terms and you get

5.406*final = -33.29 C

Divide both sides by 5.406 and you get your answer of

final temperature = -6.158 C

Gotta Love Thermo

Answer 2

After what period of time?