# Question f9cc1

Jan 13, 2016

All of the ice is melted and final temperature of the water is ${100}^{o} C$ with a small amount of steam.

#### Explanation:

1st of all, I think this is in the wrong section.
2nd, you might have misinterpreted some data which, if changed, can change the exercise's way of solving. Check the factors below:

Suppose the following:

• Pressure is atmospheric.
• The 20g at ${100}^{o} C$ is saturated steam, NOT water.
• The 60g at ${0}^{o} C$ is ice, NOT water.

(The 1st one only has minor numerical changes, whereas the 2nd and 3rd have major changes)

There are different scenarios for this. Suppose that ice melts and is converted to water. The key here is to understand that as water changes phase (solid, liquid, gas = ice, water, steam) the heat given during the transition will not increase the temperature. This heat is called the latent heat.

Ice will therefore consume some energy to melt (melting latent heat).

Steam will consume some energy to evaporate (boiling latent heat).

Therefore, if we take as a reference temperature ${T}_{0} = {0}^{o} C$ (as many engineering tables and manuals do) ice has negative heat change and steam has positive heat change . Therefore, the final energy will be equal to:

Q=ΔΗ_(s)-ΔH_i

• Ice ΔΗ_(i)

According to The Engineering Toolbox - Latent Heats of Melting, the melting latent heat is equal to $334 \frac{k J}{k g}$ But for 20 grams:

ΔΗ_(i)=334 (kJ)/(kg)*0.060kg=20.04kJ

That means that if ALL the ice is melted, then 6.68 kJ are needed.

• Steam ΔΗ_(s)

According to The Enginnering Toolbox - Steam Properties, the total heat of the steam if it boils at ${100}^{o} C$ is $2675.43 \frac{k J}{k g}$ But for 60 grams:

ΔΗ_(s)=2675.43 (kJ)/(kg)*0.020kg=53.51kJ

• Final energy

Q=ΔΗ_(s)-ΔH_i#

$Q = 53.51 - 20.04$

$Q = 33.47 k J$

For the total of $20 g r a m s + 60 g r a m s = 80 g r a m s = 0.08 k g$:

$Q = \frac{33.47}{0.08} \frac{k J}{k g} = 418.34 \frac{k J}{k g}$

• Conclusion

The table in The Enginnering Toolbox - Steam Properties says that the maximum sensible heat of water without vapor is $417.51 \frac{k J}{k g}$ Therefore, since the final heat is (even slightly) bigger, the water is boiling at ${100}^{o} C$. The only difference with the initial steam is that it was saturated (no liquid water) whereas now it is mostly water with a bit of steam.