What is the final temperature of the water when ice is added to it?

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SUMMARY

The final temperature of water when 20 g of ice at 0°C is added to 180 g of water at 22°C is calculated to be 20.86°C. The heat lost by the water is expressed as mwc(22 - t), while the heat gained by the ice incorporates the latent heat of fusion, represented as miL + mic(t-0). The calculations confirm that the latent heat of fusion is necessary for accurate results, as it accounts for the energy required for the phase change of ice to water.

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Question #1

Homework Statement


How much heat will be lost when 120 g of steam at 100°C is condensed to water at 80°C?

Homework Equations



Q = mc∆t
Q = m LV

The Attempt at a Solution



Q = mc∆t
Q = 0.12 x 4186 x (80 – 100)
Q = 10046.4 J
Q = m LV
Q = 0.12 x (- 2.3 x 106)
Q = 276000 J
= 10046.4 + 276000 = 286046.4
= -2.9 x 105 J

Is the latent heat of condensation negative to that of vaporization?

Question #2

Homework Statement


If 20 g of ice at 0° is added to 180 g of water at 22°C, what will the final temperature of the water be?

Homework Equations


Q = mc∆t

The Attempt at a Solution


mc∆t = mc∆t
0.02 x 2050 (0-T) = 0.18 x 4186 (T-22)
-41T = 753.48T - 16576.56
-794.48T = -16576.56
-794.48T/-794.48 = -16576.56/-794.48
T = 20.86

For this question, is the latent heat of fusion needed?
 
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Yes, it has the opposite sign, as energy is required for vaporization but released by condensation. You calculations look good.
 
Mapes said:
Yes, it has the opposite sign, as energy is required for vaporization but released by condensation. You calculations look good.

Thanks for the reply! I'd like someone to confirm the second question too. Thanks in advance.
 
:O no one?
 
missileblitz said:
:O no one?

Heat lost by water = mwc(22 - t)

Heat gained by the ice = miL + mic(t-0) where L is the latent heat of fusion of ice.

Now solve for t.
 
rl.bhat said:
Heat lost by water = mwc(22 - t)

Heat gained by the ice = miL + mic(t-0) where L is the latent heat of fusion of ice.

Now solve for t.

Many thanks for your reply!
 

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