How Does Adding Ice to Water Affect Final Temperature?

[aerogurl2]

I need some help on this problem

If 0.25kg of ice at 0 celsius is added to 0.15 kg of water at 20 celsius

a)Does all the ice melt? Explain
b)What is the final temperture?

Can someone please explain or show me how to get the answer to these two problems because I'm struggling to figure them out. Thanx

 

[Curious3141]

The amount of heat energy needed to convert ice from the solid state at 0 deg C to liquid water at 0 deg C is called the specific latent heat of fusion and its value is 334 kJ/kg.

The amount of heat energy liberated from the cooling of liquid water from temperature T to temperature t without phase change can be calculated from Q = mc(T-t) where m is mass and c is the specific heat capacity of water (4183 J/kg/K).

For all the ice to melt, a certain amount of heat must be supplied. Calculate this. Now find out how much heat can potentially be liberated by cooling the orig. mass of liquid water from 20 deg C to 0 deg C. Compare the two quantities to draw the conclusion for part a).

 

[aerogurl2]

If the energy of the ice is .25(2.09x10^30)= 522.5 J
and the energy of the water is .15(4.186x10^3)= 627.9 J
and the difference in the two is 105.4 J, how do I find the final temperture
I'm sort of getting confused

 

[Curious3141]

If the energy of the ice is .25(2.09x10^30)= 522.5 J
and the energy of the water is .15(4.186x10^3)= 627.9 J
and the difference in the two is 105.4 J, how do I find the final temperture
I'm sort of getting confused

In your calculation of the energy req'd to melt the ice, the latent heat of fusion is wrong (where did you get your value from). I'm assuming the 10^30 is a typo, you meant 10^3. Note that specific latent heat of fusion of ice is NOT the same thing as specific heat capacity of ice.

In your calculation of the energy liberated from cooling the water, the specific heat capacity value is right, but you didn't consider the temperature change. You have to multiply by the change in temperature (20 degrees in this case).

 

[aerogurl2]

the energy for ice would be .25(3.33x10^5)=83250 J
the energy for water would be .15(4.186x10^3)(20)= 12558 J
the difference in the energy is 70692 J ?

 

[Curious3141]

the energy for ice would be .25(3.33x10^5)=83250 J
the energy for water would be .15(4.186x10^3)(20)= 12558 J
the difference in the energy is 70692 J ?

Your values are correct. Now, since the heat energy required to melt all the ice is greater than what the liquid water can possibly supply, what's your conclusion for part a)?

From the answer to part a), there can only be one possible answer for part b) (it doesn't need any further calculation, even).

 

[aerogurl2]

for part a it's no right? since the ice's energy is graeter than the water's energy

 

[Curious3141]

for part a it's no right? since the ice's energy is graeter than the water's energy

The answer is "no", which is correct. But aerogurl, you should learn to be more precise with your language - try to phrase things like I did instead of saying simply "energy of ice" or "energy of water". It might well make a difference to most teachers, who will deduct marks for vagueness.

Have you figured out what b) should be?

 

[aerogurl2]

i don't understand how I can find the final temperture though

 

[Curious3141]

i don't understand how I can find the final temperture though

In general, when you take a water-ice mixture at zero degrees Celsius and apply heat, what will happen is that the mixture remains at zero deg C until all the ice melts. Only after all the ice has melted to water, will the temperature of the water start rising.

Now you have this system which is closed off from outside heat sources, so all you have is the water and ice. You've determined that when this closed system reaches thermal equilibrium, there is still ice left over. So what is the final temperature of the mixture?

Not meaning to confuse you, but I'd like you to figure this out step by step by yourself, so you can understand better.

 

[Curious3141]

Let me make things clearer with another example. Suppose you added a small quantity of ice at zero deg C to a large quantity of water at 20 deg C. In my example, the water will cool by (say) 10 degrees C at which point all the ice would've melted to water. Then the heat energy would redistribute between the "original" water and the "new" water formed by melting to give a final temperature slightly lower than 10 degrees Celsius.

In this scenario, however, there's "too much" ice. The water cools and cools but it cannot release enough heat to melt all the ice. In the end, the water has cooled to a certain "lowest" temperature but ice still remains. This is the final temperature - can you figure out what it should be?

 

[aerogurl2]

is it still zero degrees celsius since not all the ice have melted therefore the temperture cannot continue to rise

 

[Curious3141]

is it still zero degrees celsius since not all the ice have melted therefore the temperture cannot continue to rise

Excellent, correct!

 

[aerogurl2]

thank you so much Curious3141! for going through this problem with me step by step and having so much patience with me now i understand it and sure I'll try to be more precise with my language Thanks for the help!

 

[Curious3141]

You're most welcome.